MATHEMATICS

Q.1 lim  sin x  a tan x 

x a 2 2a 

a a  
(A) (B) – (C) (D) –
  a a
[B]

Q.2 Lim { 3 ( n  1) 2  3 ( n  1) 2 }
n 
(A) 0 (B) 1 (C) 2 (D) 3 [A]

Lim Lim
Q.3 m  n  

1  n 1n  2 n  n 2 n  3n  n 3n  4 n  ...  n (m  1) n  m n 
 
 m2 
 

is equal to
1 1 1 1
(A) (B) (C) (D) [A]
2 3 4 5
Lim
Q.4 x 0 
(– n ({x} + | [x] | )){x} is-
(Here {x} & [x] are respectively fractional part & greatest integers of x)
(A) 0 (B) e1
(C) n 2 (D) n ½ [D]
1 3 1 1
[13 x ]  [ 2 x ]  [33 x ]  ........  [ n 3 x ]
Q.6 lim 2 3 n is –
n  2 2 2
1  2  .......  n
(Where [.] denotes the greatest integer function)
2x x
(A) x (B) (C) 0 (D) [A]
3 6

Q.7 Suppose you have two linear functions f and g shown below.
f(x)
(0, 6)
(x, f(x)) g(x)
(0, 3)
(x, g(x))
(a, 0) x O
f (x)
Then Lim is -
x a g(x )
(A) does not exist
(B) not enough information
(C) 2
(D) 3 [C]
Sol.[C] This problem requires a geometrical argument :
f (x ) xa g( x ) f (x) 6
Method.1 By similar triangles, = = , and therefore = =2
6 0a 3 g ( x ) 3
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 f (x )
Method.2 lim
f (x)
= lim
 a = lim slope of f = 6 = 2
x a g ( x ) x a g( x ) x a slope of g 3
a
This problem is a nice preview of L'Hospital's Rule

1/ x
 f (1  x ) 
Q.8 Let f: R  R be such that f(1) = 3 and f '(1) = 6, Then, lim   is equal to
x 0  f (1) 

(A) 1 (B) e1/2

2
(C) e (D) e3 [C]
1/ x
 f (1  x ) 
Sol. lim  
x 0  f (1) 

1/ x
 f (1  x )  f (1) 
= lim 1  
x 0  f (1) 
f ( x 1) f (1)
lim
= x 0 x f (1)
e
1 f ( x 1) f (1)
lim
= f (1) x 0 (x)
e
f '(1)  f (1  x )  f (1) 
= f (1) xlim  f ' (1) 
e  0 x 
= e6/3 = e2

4 3 2
Q.9 If xlim{( x  ax  3x  bx  2

 x 4  2 x 3  cx 2  3x  d ) }

is finite, then the value of a is

(A) 3 (B) 5
(C) 2 (D) any real number
[C]

Sol. lim ( x 4  ax 3  3x 2  bx  2 )
x 

 x 4  2 x 3  cx 2  3x  d )

 4 3 2
 ( x  ax  3x  bx  2)
= lim 
x   4 3 2
 x  ax  3x  bx  2

 ( x 4  2 x 3  cx 2  3x  d ) 


 x  2 x  cx  3x  d 
4 3 2


 3
 (a  2) x  (3  c) x
2
lim 
x   4 3 2
 x  ax  3x  bx  2

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 
 ( b  3) x  ( 2  d )


 x  2 x  cx  3x  d 
4 3 2

Clearly, the degree of the polynomial in the
numerator is 3 and that of denominator is 2.
Therefore, for the limit to be finite, we must have,
a–2=0a=2

1 5 
Q.10 If f(x) = f ( x  1)   and f(x) > 0 for all x  R, then xlim

f(x) is
3 f ( x  2) 

2 5
(A) (B)
5 2
(C)  (D) does not exist [B]

Sol. Let xlim


f(x) = I then,

lim f ( x  1) lim f ( x  2)  I
x  x 

1 5 
Since, f ( x )  f ( x  1)  
3 f ( x  2) 
 
1 5 
 xlim

f (x)   lim f ( x  1)  
3  x  lim f ( x  2) 
 x  
1 5
 I =  I    3I2 = I2 + 5
3 I
5
 2I2 = 5  I= 
2

100 x   99 sin x  
Q.11 The value of lim   
x 0  x   x  
 sin
where [ ] represents the greatest function is-
(A) 199 (B) 198 (C) 0 (D) 197 [B]
Sol. We know that

x > sin x for all x > 0

and x < sin x for all x < 0

x sin x
 > 1 and < 1 for x  0
sin x x

100 x 99 sin x
 > 100 and < 99
sin x x

 100x   99 sin x  
 lim     
x 0  sin x 
  x 

= xlim
0
(100 +98) = 198

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 
 12 3 52 7 

Q.12 The lim  3
 2
 3
 2
 .... equals -
x  1  x 1  x 1  x 1  x 
 
5 10 5 21
(A)  (B)  (C) (D) [B]
6 3 6 3

 1
2
3 52 7 

Sol. lim  3
 2
 3
 2
 ....
x  1  x 1  x 1  x 1  x 
 

1  5  9  .... 3  7  11  ..... 
2 2 2

= lim  3
 2 
x   1 x 1 x 
 
 x x 


 (4k  3)
k 1
2
 (4k  1) 
k 1
= xlim   

 1 x3 1 x2 
 
 
 x x x x x 
16
 k 1

k 2  24
k 1
k  9
k 1
4  k  1 
k 1 k 1
lim
= x    
 1 x3 1 x2 
 
 

 16 x ( x  1)( 2 x  1)
  12x ( x  1)  9 x
 6
= lim 
x 
 1 x3



2 x ( x  1)  x 
 
1 x2 

32 10
= – +2=  .
6 3

Q.13 Let the sequence < bn> of real numbers satisfy the recurrence relation :

1  125 
b n 1  2b n  2  , bn 0, then nlim b n is equal to -
3  b n  

2
(A) 0 (B)  (C) 5 (D) [C]
3

Sol. Let nlim b =b

 n

 
Then, bn +1 =
1  2b n  125 
3  2 
bn 

 
lim 1 125 
 2 lim b 
n  bn + 1 = 3  n  n 2 
 lim b n 
 n  
1  125 
 b=  2b  2 
3  b 

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 b 125
 = 2  b3 = 125  b = 5
3 b

sin x  (sin x ) sin x

Q.14 The value of lim is-
x   / 2 1  sin x  n sin x

(A) 0 (B) 1
(C) 2 (D) None of these [C]
Sol. Let sinx = t

if x  , t1
2

t  (t) t
so lim
t 1 1  t  nt

using L Hospitals

1  t t (1  log t )
lim
= t 1 1
1
t
1
 t t (1  log t ) 2  t t  
= lim  t  = +2
t 1 1
 2
t

xe x  n (1  x )
Q.15 lim is equal to-
x 0 x2
3 2
(A) (B) (C) 0 (D) 1 [A]
2 3
Sol. Use expansion.
e1/ x  e 1/ x
Q.16 Let f(x) = g(x) and x 0 where g is a continuous function. Then xlim
0
f(x) exists if
e1/ x  e 1/ x
(A) g(x) is any polynomial
(B) g(x) = x + 4
(C) g(x) = x2
(D) g(x) = 2 + 3x + 4x2
e1 / x  e 1 / x 1  e 2 / x
Sol.[C] lim = lim = 1 and
x 0 e
1/ x
 e 1 / x x 0  1  e
2 / x

e1 / x  e 1 / x e2 / x  1
lim 1 / x = lim = –1. Hence
x 0 e  e 1 / x x  0  e 2 / x  1
lim f ( x ) exists if g(x) = x or g(x) = x2.
x 0

If g(x) = a (a  0), then xlim

0
f ( x ) = a and lim f ( x ) = –a. Thus lim f ( x ) doesn't exist in this case.
x 0  x 0

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   
Q.17 If fr() =  cos 2
 i sin ×
 r r2 

 2 2    
 cos 2  i sin 2  ….  cos  i sin 
 r r   r r

then nlim

f n () equals

(A) –1 (B) 1 (C) – i (D) i

Sol.[D] Using De Moivre's theorem
= e i / r e 2i / r …. e i / r
2 2
fr ()
2
= e ( i / r )(1 2 ...... r )

= e ( i / 2)(11 / r )
2
= e ( i / r )[ r ( r 1) / 2 ]

 nlim f n ( ) = lim e i / 2(11/ n )

 n 

 
= ei/2 = cos   + i sin   = i
2 2

Q.18 Let

f(x)= nlim
 {sinx + 2sin x + 3sin x +...+ n sin x}. If x  n +
2 3 n
, n  I, then
2
1
lim 2 is equal to :
x  / 2 [(1 – sin x ) f ( x )] x –1
sin

(A) 1 (B) 0
(C) e (D) e2
sin x
Sol.[C] f(x) =
(1 – sin x ) 2
1
lim
Now x  2 sin x –1
 / 2 [(1 – sin x ) f ( x )]

1
= lim
x  / 2 (sin x ) sin x –1
sin x –1
= lim =e
e x / 2 sin x –1

(2sin x  1)[n (1  sin 2x )]

Q.19 lim is equal to
x 0 x tan 1 x
(A) n 2 (B) 2 n 2
2
(C) (n 2) (D) 0

( 2 sin x  1)[n (1  sin 2 x )]

lim
Sol.[B] x 0 tan 1 x
x2
x

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 2 sin x  1 sin x n (1  sin 2 x )
= lim × ×
x  0 sin x x sin 2 x

sin 2 x
× × 2 = 2n2
2x
1

Q.20 If f(n) = Lim  x  x   x  x

x0  1  sin  1  sin  ..... 1  sin n  
2  2
 2   2 

then nLim
  f(n) =

(A) 1 (B) e (C) 0 (D) 

1  x  x   x  
  1 sin   1 sin 2 .... 1 sin n  1
Sol.[B] f(n) = Lim e x   2  2   2  
x 0

  x x x   x x  
 1  sin  sin  ......  sin    sin sin 2 ...... .....1
2 2 n 2
  2 2   2  
=
Lim e x
x 0

  x   x 
 sin x sin  2   
 2   2   ....... sin  2 n  
 x x 
= 
x

Lim e 
x 0

1 1 1 
=   2 ..... n 
 2 2 2 
e
1/ 2
Sum of infinite G.P. = 1 =e
1
e 2

Q.21 lim | x  1 |  | x – 1 | –2 equals -

x 0
x
(A) 1 (B) –1 (C) 2 (D) 0 [D]

Q.22 lim [cot–1x] is; where [ ] represents greatest integer function -

x – 


(A) 0 (B)
2
(C) 3 (D) None of these [C]

Q.23 If f(x) = x2 + bx ; 0  x  1
= 3 – ax2; 1 < x  2
such that lim
x 1
f(x) = 4, then ordered pair (a, b) is given as-
(A) (1, 5) (B) (1, 7)
(C) (–1, 3) (D) None of these [C]

1 x 2 1 1
Q.24 lim
x 0 x 5 0 e t dt –
x 4

3x 2
is equal to

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 (A) 1 (B) 3/5 (C) 1/2 (D) 3/10
Sol.[D] Applying L'Hospital Rule
x
t2
3 0 e dt  3x  x 3  0 
 
x5
0
2
lim 3e  x  3  3x 2
x 0
5x 4
2
=
3 lim e
x
 1  x 2  0 
x 0
5 x4 0
2
3 lim e  x (2 x )  0  2 x
=
5 x 0 4x 3
2
3 2 lim  e  x  1  0 
= ·  
5 4 x 0 x2 0
2
3 2 lim 1  e  x 3
=  x  0 2
=
5 4 (x ) 10

Q.25 A particle begins at the origin and moves 2 units to right and reaches P 1 then 1 unit up and reaches P2, 1/2 unit right
1
and reaches P3, unit down to reach P4 & 1/8 unit right to reach
4
P5 and so on. If Pn = (xn, yn) then nlim P is
 n
(A) (4, 6) (B) (8/3, 4/5)
(C) (4/5, 2) (D) (4, 3)
  1 n 
21    
 4 
Sol.[B] nlim
 n
x =     = 8/3
1
1
4
   1  
n

1 1  
lim yn =   4  
= 4/5
n 
 1 
1  
 4 
(8/3, 4/5)
2
Q.26 lim f ( x ) , where 2 x  3 < f(x) < 2 x  5x , is
x 
x x2
(A) 1 (B) 2 (C) –1 (D) –2 [B]

 4 2 2 
Q.27 If lim  x  x  1  ax  b   0 then
x   
(A) a = 1, b = – 2 (B) a = 1, b = 1
1
(C) a = 1, b =  (D) None of these [C]
2

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 Lim
Q.28  [1 + (cos x)cos x]2 is equal to
x
2
(A) Does not exist (B) 1
(C) e (D) 4
Lim
Sol.[D] x    [1 + (cos x)cosx]2
2

Lim
y=  (cos x)cos x
x
2
Lim
log(y) =  (cos x) log cos x
x
2
Lim log(cos x )
log (y) =  (/) L'hospital
x sec( x )
2
Lim 1 sin x
log(y) =  ×–
x cos x sec x tan x
2
Lim
=  – cos x = 0
x
2
y = e0 = 1
Now limit is (1 + 1)2 = 22 = 4

2 2  (cos( x )  sin( x ))3 Lim

Q.29 Let f(x) = then x   f(x) is equal to
1  sin( 2 x ) 4
1 3
(A) (B) 2 (C) 1 (D)
2 2

2 2  (cos x  sin x )3
Sol.[D] f(x) = (0/0)
1  sin 2 x
L'Hospital Rule
lim
 f(x)
x
4
lim 3(cos x  sin x ) 2 (  sin x  cos x )
= x  –
4  2 cos 2 x
lim  3(cos x  sin x ) 2 (cos x  sin x )
= x 
4 2(cos x  sin x ) (cos x  sin x )
lim
3
= x  × (cos x + sin x)
4 2
3 2 3
=  =
2 2 2

S3n 2n
f (r )
Q.30 If Sn denote the sum of first n terms of an A.P. and f(n) =
S2 n  S n
, then nLim
  n
is equal to
r 1
(A) 6 (B) 3 (C) 2 (D) 0
3n
Sol.[A] S3n = [2a + (3n – 1)d]
2

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 2n
S2n = [2a + (2n – 1) d]
2
n
Sn = [2a + (n – 1)d]
2
n
S2n – Sn = [2a + (3n – 1)d]
2
S3n
= 3 = f(n)
S2 n  S n
2n
1  2n 
n
S = xLim

f (r ) = nLim

3  =6
r 1  n 

Q.31 If  and  are the roots of the quadratic equation ax2 + bx + c = 0, then

Lim 1  cos(cx 2  bx  a )
x 1 / 
=
2(1  x ) 2

c  1 c  1
(A)    (B)   
2    

 2    



c   1
(C)    (D) None of these
    

Sol.[A] ax2 + bx + c = 0 roots , 

So, cx2 + bx + a = c (x – 1/) (x – 1/)

Lim 1  cos(c( x  1 /  )( x  1 / ))

x 1 /  2 2 ( x  1 / ) 2

c
Lim 2 sin 2 ( x  1 / )( x  1 / )
x 1 /  2
2 2 ( x  1 /  ) 2

 2 c 
 sin  ( x  1 / )(x  1)   2
Lim 1   2  2 c
 ( x  1 / )
2   2 
x 1 / 
4
 ( x  1 /  )(x  1 / ) c  
 2  
 

Lim 1 c2 c 1 1
x 1 /   (1 /   1 / ) 2 = 2     
2 4  

Q.32 Lim log sin( x / 2) sin x is equal to -

x 0 

(A) 1 (B) 0 (C) 4 (D) 1/4

log sin x cot x
Sol.[A] Let y = lim  x  = xlim  2 x
x 0 log sin   0 cot
2 2
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 tan x / 2
= lim 2
x/2 × x/2 =1
x 0 tan x x
x

Q.33 lim  1  2  ......  n  is equal to:

n 
1 n 2 1 n2 1 n2 
(A) 0 (B) –1/2
(C) 1/2 (D) None of these

 1 2 n 
Sol.[B]  2
 2
 ......  2
1 n 1 n 1 n 
1
= × (1 + 2 + ..... + n)
1 n 2
1 n (n  1) n 1
= . = =
1 n 2
2 2(1  n ) 2 [(1 / n )  1]
 1 2 n 
 nlim

 2
 2
 ......  
1 n 1 n 1 n2 
1 1
= nlim
 2 [(1 / n )  1]
=–
2

 [ x ]2  sin[ x ]
 for [ x ]  0
Q.34 If f(x) =  [x]
 0 for [ x ]  0

where [x] denotes the greatest integer less than or equal to x, then xlim
0
f(x) equals:

(A) 1 (B) 0
(C) –1 (D) None of these
Sol.[D] As x  0 – (i.e., approaches 0 from the left),
[x] = –1,

 lim f(x) = lim 1  sin (1) = –1 + sin 1

x 0  x 0 1
whereas, if x  0+ we get [x] = 0,

f(x) = 0  lim f(x) = 0

x 0 

Thus, xlim
0
f(x) does not exist.

lim 1  cos 3 x
Q.35 x 0
is equal to
x sin x cos x
(A) 2/5 (B) 3/5 (C) 3/2 (D) 3/4

1  cos 3 x
Sol.[C] xlim
0
= lim
x sin x cos x x 0

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 (1  cos x ) (1  cos 2 x  cos x )
x sin x cos x

2 sin 2 ( x / 2) (1  cos 2 x  cos x )

= xlim
0 2 x sin ( x / 2) cos ( x / 2) cos ( x / 2) cos x
1 sin ( x / 2) 1  cos 2 x  cos x
= xlim
0
. .
2 x/2 cos ( x / 2) cos x
1 3 3
= . 1. =
2 1 2

1 1
Q.36 lim sin x  tan x is equal to
x 0
x3
(A) 2 (B) 1 (C) –1 (D) 1/2

sin 1 x  tan 1 x 0 
Sol.[D] xlim
0
 form 
x 3 0 
1 1

= xlim
0 1 x 2 1  x 2 (L Hospital rule)
3x 2
1 lim 1 1  x 2  1  x 2 
=  
3 x 0 x 2 (1  x 2 ) 1  x 2 
 
1 lim 1  (1  x 2 ) 2  (1  x 2 ) 1 
=  . 
3 x 0 x 2  (1  x 2 ) 1  x 2 (1  x )  1  x 2
2 

1 lim x 2 (3  x 2 ) 1
= .
3 x 0 x 2 (1  x 2 ) 1  x 2 (1  x 2 )  1  x 2
1 3 1
=  =
3 2 2

Q.37 Given f(2) = 6 and f(1) = 4

2
lim f (2h  2  h )  f ( 2) is equal to
h 0 2
f (h  h  1)  f (1)
(A) 3/2 (B) 3 (C) 5/2 (D) –3

f (2h  2  h 2 )  f ( 2)
Sol.[B] hlim
0 f (h  h 2  1)  f (1)

f ( 2h  2  h 2 )  f (2) h ( 2  h )
= hlim ×
0
2h  2  h 2  2 h (1  h )

(h  h 2  1)  1
×
f (h  h 2  1)  f (1)

2h 1 1
= f(2) × hlim
0 × =6×2× =3
1 h f ' (1) 4

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 Note that L Hospital rule is not applicable in this case.

|x|
Q.38 The value of lim is -
x0 x

(A) 1 (B) 2
(C) 3 (D) Does not exist [D]

 2
x  1 , x  1
Q.39 If f(x) =  then the value of lim
x 1
f(x) is -
3x  1 , x  1

(A) 1 (B) 2
(C) 3 (D) Does not exist [B]

Q.40 Lim (1 – x + [ x – 1] + [1 – x]) where [x] denotes greatest integer but not greater than x
x 1

(A) 1 (B) –1
(C) 0 (D) Does not exist [B]

1/ x
Lim e 1
Q.41 x 0
=
1/ x
e 1
(A) –1 (B) 1
(C) 0 (D) Does not exist [D]

1
Q.42 If f ( x )  3  1 /(1 x )
then-
1 7
(A) Lim f(x) = 3
x 1

(B) Lim f(x) = 4

x 1

(C) Lim
x 1
f(x) = 4

(D) Lim
x 1
f(x) does not exist [D]

Q.43 lim[cos 1 (cos x )] , where [] denotes greatest integer function

x 1

(A) 0 (B) 1
(C) Does not exist (D) None of these [C]

  cos –1 x 0
Q.44 lim  ,   =
x  1 x 1 0
1
(A)  (B)
2
(C) 2 (D)  [C]

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 cos θ  sinθ
lim
Q.45 π π =
x 
4 θ
4
(A) 2 (B) 1
(C) 2 (D) Does not exist [A]

x3  x 2 1
Q.46 lim =
x  x3  x2 1
(A) 0 (B) 1
(C) 2 (D) Does not exist [B]

x4  x2 1
Q.47 lim =
x  x5  x 2 1
(A) 0 (B) 1
(C) 2 (D) Does not exist [A]

3x 5  x 2  13
Q.48 lim =
x  x 4  7 x 2  17
(A) 0 (B) 2
(C) infinite (D) None [C]

x 3  2x  1
Q.49 lim =
x  1 x 5  2x  1
(A) 2/3 (B) 1/3 (C) 4/3 (D) 5/3 [B]

2 sin 2 x  sin x  1
Q.50 lim =
x
 2 sin 2 x  3 sin x  1
6

(A) 0 (B) 3 (C) –3 (D) 1 [C]

sin 3 x  x 3
Q.51 lim =
x  0 (sin x  x )

(A) 0 (B) 1 (C) –1 (D) 2 [A]

x 3  7x 2  15x  9
Q.52 lim =
x 4  5x 3  27x  27
x 3

2 2 1
(A) (B) (C) (D) 1 [B]
3 9 9

m
x 1
Q.53 lim , (m  n) ; m, n  I
x 0 n x 1
(A) mn (B) m/n
(C) n/m (D) None [C]

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Q.54 lim  x 2  1  x 2  1  =
x  0 
(A) 0 (B) 1
(C) 2 (D) None [A]

(1  x 2 )  (1  x )
Q.55 lim =
x 0 3
(1  x )  (1  x )
(A) 0 (B) 1 (C) 2 (D) 4 [B]

x  2a  x 2a
Q.56 lim , a>0=
x  2a  2 2
x  4a
(A) 2a (B) 2 a
(C) 1 / 2 a (D) a [C]

2 x  sin 1 x
Q.57 lim =
x 0 2 x  tan 1 x
1
(A) 3 (B) (C) 0 (D) 1 [B]
3

1 1
Q.58 lim  2 log (1 + x) =
x 0 x x
1
(A) 1 (B) (C) 0 (D) 2 [B]
2

sin x
Lim e  1  sin x
Q.59 x 0
=
x2
1
(A) 1 (B) (C) e1/2 (D) e [B]
2

log e {1  tan( x  a )}
Q.60 lim =
x a tan( x  a )
(A) 0 (B) 1 (C) 2 (D) 3 [B]

cos(x / 2)
Q.61 lim =
x 1 1 x
(A) 0 (B)  (C) /2 (D) 2 [C]

(cos ) x  (sin ) x  1
Q.62 lim , x  (0, /2)
x 2 x2
(A) sin2 n (sin )
(B) cos2  n (cos )
(C) cos2  n (cos ) – sin2 n (sin )
(D) cos2  n (cos ) + sin2 n (sin ) [D]

Q.63 lim (1  k / x ) mx =
x 
(A) ek (B) e
(C) emk (D) None [C]

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 x
 x 
Q.64 lim   =
x   1  x 

1
(A) e (B) (C) 0 (D) 1 [B]
e

Q.65 lim (1  x )1/(13 x ) =

x 0
1
(A) e13 (B)
e 13
(C) e (D) 1 [B]

log e (a  x )  log e a
Q.66 lim =
x 0 x
(A) a (B) a2
(C) a–1 (D) Does not exist [C]

log e (1  7 sin x )
Q.67 lim =
x 0 sin x
(A) 3 (B) 7
(C) 0 (D) Does not exist [B]

tan x  sin x
Q.68 lim =
x 0 x3
1 1
(A) (B)
2 4
(C) e (D) Does not exist [A]

tan x
Lim e  ex
Q.69 x 0
=
tan x  x
(A) e (B) 1
(C) 0 (D) Does not exist [B]

o
Q.70 Lim sin x is equal to
x 0
x
(A) 0 (B) 

(C) 1 (D) [D]
180

1 x

Q.71  1  x  1 x 2 =
lim 
x 1 2  x 

(A) (2)1/3 (B) (2/3)1/2

(C) (2/3)1/4 (D) Does not exist [C]

Q.72 lim x1 / x =
x 
(A) 0 (B) 1
(C) 2 (D) (2)1/2 [B]

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 1/ x3
Q.73  tan x  =
lim  
x 0 x 
(A) 0 (B) 1
(C) e (D) Does not exist [D]

x
Q.74 Lim tan
=
x 1 (2  x ) 2

(A) e–2/ B) e1/

(C) e2/ D) e–1/ [C]

Q.75 Lim sin x  log(1  x ) =

x 0
x2
(A) 0 (B) 1/2
(C) –1/2 (D) does not exist [C]

Q.76 Lim (1  cos 2 x ) sin 5x =

x 0
x 2 sin 3x
(A) 10/3 (B) 3/10 (C) 6/5 (D) 5/6 [C]
x2
xe
Q.77 Lim x =

2
x 0 e t dt
0
(A) 0 (B) 1
(C) –1 (D) does not exist [B]
Q.78 Let f(a) = g(a) = k and their nth derivatives exist and are not equal for some n.
Further if
lim f (a )g ( x )  f (a )  g (a )f ( x )  g (a )  4
x a g( x )  f ( x )
then k is equal to
(A) 0 (B) 4 (C) 2 (D) 1 [B]

x2  x  1
Q.79 lim  (where [x] is greatest integer function  x)
x  e[ x ]
(A) 0 (B) 1
(C) 2 (D) Does not exist [A]

Q.80 lim [ x ]  [ 2 x ]  [3x ]  .....  [ nx ] =

n  1  2  3  ......  n

(where [·] denotes the greatest integer function)

x x
(A) 0 (B) (C) (D) x [D]
2 6

1 1 1
Q.81 lim [ x ]  2 [ 2 x ]  3 [3x ]  ...  n [nx ] =
n 
12  2 2  32  ...  n 2

(where [·] denotes the greatest integer function)

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 1 1
(A) 0 (B) (C) (D) 1 [A]
2 6

x  x 2  x 3  ........x 100  100

Q.82 lim =
x 1 x 1
(A) 502 (B) 50 × 101
(C) 50 × 100 (D) 100 × 101 [A]

lim
Q.83  [Max (sinx, cosx)]
x
4

1
(A) (B) 0
2
(C) – 1 (D) 1 / 2 [D]

Q.84 Lim (cosec x)1/log x =

x 0 

(A) e (B) e–1

(C) e2 (D) 1 [B]

Q.85 Which of the following statement is/are

correct -
Lim
(A) x    [sgn sin x] = 1
Lim
(B) x    [sgn sin x]  –1
(C) Lim
x   [sgn sin x] = 1
Lim
(D) x [sgn sin x] does not exist
(Where [.] represent greatest integer function) [A]

Q.86 Let rth term of a series given by

n
r
tr 
1  3r 2  r 4
. Then Lim
n 
t
r 1
r is

(A) 3/2 (B) 1/2

(C) – 1/2 (D) – 3/2 [C]

Q.87 The sum to infinity of the series :

3 5 7
3
+ 3 3
+ + ...... is -
1 1 2 1  2 3  33
3

(A) 3 (B) 4 (C) 5 (D) 6 [B]

Q.88 Let
     2 2    
fk () =  cos 2
 i sin   cos 2  i sin 2  .... ….  cos  i sin  then Lim f () =
n  n
 k k2   k k   k k

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  
(A) cos + i sin (B) cos + i sin
2 2
 
(C) i cos + sin (D) i cos + sin [A]
2 2

sin [ x  ]
Q.89 Let f(x) = where [x] stands for the greatest integer function. Then xLim
n
f(x) = ?
1  [ x ]2
(Here n  I) = ?
(A) Does not exist (B) Equals to 0
(C) Equals to 1 (D) Equals to –1 [B]

 2 
Q.90 If nlim  an  1  n  = b, a finite number then the ordered pair (a, b) is -
   1 n 

(A) (1, 1) (B) (–1, 1)
(C) (1, –1) (D) None of these [A]

3 5 7
Q.91 Let Sn= + + + ... upto n terms, then lim Sn is -
3 3 3 n 
1 1 2 1  2 3  33
3
(A) 2 (B) 3
(C) 4 (D) none of these [B]

100

Q.92  (n  r) 10
=
r 1
Lim
n  n10  10 10

(A) 100 (B) 200

(C) 300 (D) None of these [A]

k
Q.93 Lim n cos n ! 0 < k < 1
n
n 1
(A) 0 (B) 1!
(C) 2! (D) None of these [A]

Sn 1  Sn
Q.94 If Sn = a1 + a2 + ........ an and Lim an = a, then Lim n is equal to-
n  n 
k
k 1
(A) 0 (B) a (C) 2 a (D) 2a [A]

 1
Q.95 The sum  is equal to -
n 1 n ( n  1)( n  2)
1 1 1
(A) 1 (B) (C) (D) [C]
2 4 8

Q.96 The continued product of

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  1  1  1   1 
 1   1    1   ........... 1  2  is Pn ;
 4  9  16   n 
then nLim P is-
 n
(where n N)
1 n 1
(A) – (B)
2 n
1
(C) (D) None of these. [C]
2

Q.97 Inscribed in a circle of radius R is a square, a circle is inscribed in the square, a new square in the circle, and so on
for n times. Find the limit of the sum of areas of all the circles and the limit of the sum of areas of all the squares as
n 
(A) 2R2, R2 (B) R2, 4R2
2
(C) 2R , 4R 2 (D) 4R2, 2R2 [C]

lim x sin( x  [ x ])
Q.98 x 1 , where [·] denotes the greatest integer function, is equal to -
x 1
(A) 1 (B) –1
(C)  (D) Does not exist [D]

Q.99 The value of x Lim 1  sin 2 x =

 / 4
  4x
1 1
(A) – (B)
4 4
1
(C) (D) None of these [D]
2

Q.100 Evaluate :
Lim [ x ]  [ x 2 ]  [ x 3 ]  ......  [ x 2 n 1 ]  n  1
x 0 
1  [ x ] | x | 2 x

(A) 1 (B) 0
(C) 2 (D) None of these [B]

Q.101 Lim cos x  1  cos x  =

x 
(A) 0 (B) 1
(C) 2 (C) None of these. [A]

 1   1   
Q.102 Lim 1  1  1   1 
1
 ...... … 1  n  is equal to-
n   2    
5  5   54   52 
5 4 1
(A) 0 (B) (C) (D) [B]
4 5 5

3
Lim x2 1  x2 1
Q.103 x  4 5
x4 1  x4 1
(A) 0 (B) 1
(C) 2 (D) None of these [B]
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Q.104 If the rth term, tr, of a series is given by
n
r
n   r
tr = . Then lim t is –
r4  r2 1 r 1
(A) 1 (B) 1/2
(C) 1/3 (D) None of these [B]

Q.105 lim
x  x2  x 1 – x2 1 =
2
(A) (B) 1
3
1
(C) (D) None [C]
2

Q.106 The value of

1/ x
lim  e xn ( 2 –1) – (2 x – 1) x sin x 
x

=
x 0  e xnx 
 
1
(A) e (B) ln 2
e
(C) e ln 2 (D) None of these. [B]

Q.107 Let f(x) = nLim


{sinx + 2sin2x + 3sin3x +...
….+ nsinn x} If sinx  n+ /2, n  :

 
1
Evaluate : Lim 1  sin x  2 f ( x ) sin x 1 =
x  / 2
(A) 1 (B) 0
(C) e (D) None of these [C]

Q.108 Lim cos(sin x )  cos x =

x 0
x4
1 1
(A) 1 (B) 6 (C) – (D) [D]
6 6

Q.109 Let a = min{x2 + 2x + 3, x  R} and

n
1  cos 
b = lim
0
 2
. Then  a r . b n r =
r 0
(A) 2n+1 + 1 (B) 2n+1 – 1
4 n 1  1 4 n 1  1
(C) (D) [C]
3 .2 n 3.2 n

Lim  ax 
Q.110 – a2  x2 cot   is-

x a
2 ax 
a 2a a 4a
(A) (B) (C) – (D) [D]
   

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  1n 
Q.111 lim n  n  e.1  1   =
n   n 
  
3
(A) 1 (B)
2
2
(C) (D) None of these [B]
3

 1  1 
Q.112 The value of Lim 1 
x 0 
 
x   tan x  4  2 
 2  
is-
(A) loga16 (B) Does not exist
(C) 3 ln 2 (D) 4 ln 2 [D]

Q.113 lim n x  [ x ] = ([.] G. I. F. )

x  [x]
(A) 0 (B) –1
(C)  (D) None of these [B]

 sin [ x ]
 , [x]  0
Q.114 If f(x) =  [ x ] [IIT 1985]

 0, [x]  0

Where [x] denotes the greatest integer less than or equal to x, then xlim
0
f(x) equals-
(A) 1 (B) 0
(C) –1 (D) None of these [D]

1
Q.115 The value of xlim
(1  cos 2 x ) [IIT-1991]
0 2
x
(A) 1 (B) –1
(C) 0 (D) None of these [D]

1/ x2
Q.116 Lim  1  5x 2 
x 0  2 
 1  3x 
(A) e2 (B) e (C) e–2 (D) e–1 [A]

Lim 1  cos 2( x  1)
Q.117 x 1
= [IIT-98, 91]
x 1
(A) Does not exist because LHL  RHL
(B) Exists and is equals to – 2
(C) Does not exist because x – 1  0
(D) Exists and is equals to 2 [A]

Q.118 Lim x tan 2 x  2 x tan x is [IIT 99]

x 0 (1  cos 2 x ) 2

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 1
(A) (B) –2
2
1
(C) 2 (D) – [A]
2
x
 x 3 
Q.119 For x  R, Lim
x    = [IIT Scr. 2000]
 x2
(A) e (B) e–1 (C) e–5 (D) e5 [C]

2
Q.120 Lim sin(  cos x) =[IIT Scr. 2001]
x 0
x2

(A) – (B)  (C) (D) 1 [B]
2
(cos x  1)(cos x  e x )
Q.121 The value of Integer n ; for which Lim
x 0
is a finite non zero
xn
number- [IIT Scr. 2002]
(A) 1 (B) 2 (C) 3 (D) 4 [C]

(sin nx ) (a – n )nx – tan x 

Q.122 If xlim
0
 0 then the value of a is –
x2
1 n
(A) (B)
n 1 n 1
1
(C) n+ (D) n [C]
n

Q.123 If f(x) is a differentiable function and f  (2) = 6,

2
f ( 2  2 h  h )  f ( 2)
f (1) = 4, f (c) represents the differentiation of f(x) at x = c, then hlim
0
f (1  h 2  h )  f (1)
[IIT Scr.2003]
(A) may exist (B) will not exist
(C) is equal to 3 (D) is equal to –3 [C]

 1
sin x 
Q.124 lim   sin x  x   1  

x 0  x  , for x > 0
 
[IIT 2006]
(A) 0 (B) –1 (C) 2 (D) 1 [D]

Questions Add (24–6-09)

 3 1
 x  2x 
1 
x  4 x  2
Lim  3   
Q.5
x 2  x  8   x 2  x 2  
   

1 1 1 1
(A) (B) (C) (D) [A]
2 4 8 16
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 n
 a 1  n b 
Q.3 Lim  ; (a > 0, b > 0) has the value equal to :
n   a 
 
(A) a1/b (B) ba (C) ab (D) b1/a [D]

0.999
  ...
9
Q.4 A function of an integral argument attains the values u1 = 0.9, u2 = 0.99, u3 = 0.999… un = . What is
n times

the Lim u equal to ? What must the value of n be for the absolute value of difference between u n and its limit not
n  n
to exceed 0.0001 ?
(A) 1, n  2 (B) 1, n  4
(C) 1, n  6 (D) 1, n  8 [B]

Q.1 If Lim
x 0
sin–1(sec x) = p and Lim
x 0
sin–1[sec x] = q where [x] denotes greatest integer function then
(A) p exists but q does not
(B) p does not exist but q does
(C) both p & q exist
(D) neither p nor q exist [B]

Q.8 If
 n2 n4 n6 5 
L = nlim

  2   ......  
 n  n  1 n  2n  4 n 2  3n  9
2
7n 
then eL is equal to -

(A) 1 (B) 1/7 (C) 7 (D) e

Sol.[C]

 n2 n4 n  2( 2 n ) 
L = nlim  
  n 2  n  1 n 2  2n  2 2
 ....  2 

 n  ( 2n ) n  ( 2 n ) 2 
2n
n  2r
= nlim
 n
r 1
2
 rn  r 2
2n
1 1  2( r / n )
= nlim

n
 (r / n )
r 1
2
 (r / n )  1
2

= x
1  2x
2
 x 1

dx = n ( x 2  x  1)  2
0
= n 7
0

 eL = 7

5n 1  3n  32n
Q.4 If nlim

is equal to
5n  2 n  32n 3
(A) 5 (B) 3
(C) 1 (D) None of these

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 5 n 1  3 n  3 2n
Sol.[D] lim
n  5 n  2 n  3 2 n 3

5.5 n  3 n  9 n
lim
n  5 n  2 n  27.9 n
n n
5 3
5.      1
 9 9 1
lim =
n n
h   5  2 27
      27
9 9

f (x 2 )  f (x)
Q.4 If f(x) is differentiable and strictly increasing function then the value of lim is
x  0 f ( x )  f (0)

(A) 1 (B) 0 (C) –1 (D) 2

f (x 2 )  f (x)  0 
Sol.[C] xlim
0
 
f ( x )  f ( 0)  0 
Apply L'Hospital rule

lim f ' ( x 2 )( 2 x )  f ' ( x )

x 0 f ' (x)
f '(x) is strictly increasing function so that f '(x) > 0
f ' (0)( 2  0)  f ' (0)

f ' (0)
f ' ( 0)
– =–1
f ' ( 0)

3 tan 1 x  3 tan x  x 5  6 x
Q.3 If lim is a finite number then the greatest value of n is
x 0 3x n
(A) 3 (B) 5
(C) 2 (D) None of these
Sol.[D] xlim
0

 3 3 5   3 2 5  5
 3x  x  x  ...   3x  x  x  ...  6x  x 
 5   5  
n
x
all coff. of x, x3 and x5 is zero and minimum power of x occurs in numerator is 7
So n=7
n
r3  8
n  
Q.4 Lim is equal to
r 3 r3  8
(A) 2/7 (B) 3/7
(C) 7/2 (D) 7/3

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 n
r3  8  33  8   43  8 
Sol. [A] Lim
n   3
r 8
=  3
3  8


 
 43  8 
r 3    

 n3  8 
........  3 

 n 8

 (3  2) (32  2 2  3.2) 
= Lim
n 
 2 2 
 (3  2) (3  2  3.2) 
 ( 4  2) ( 4 2  2 2  4.2)   (n  2) (n 2  2 2  n.2) 
 2 2  .........  2 2 
 ( 4  2) ( 4  2  4.2)   (n  2) (n  2  n.2) 

 (3  2) ( 4  2).......(n  2) 
= Lim
n   (3  2) ( 4  2).......n  2 
 

 (3 2  2 2  3.2) (4 2  2 2  4.2)......(n 2  2 2  n.2) 

 2 2 2 2 2 2 
 (3  2  3.2) (4  2  4.2).....(n  2  n.2) 

1.2.3.4.5.6........   19.28.39.52.63.......... 
=    7.12.19.28.39.52.......... 
 5.6.7.8.....   
1.2.3.4
= = 2/7
7.12

Q.5 ABC is an isosceles triangle described in a circle of radius r. If AB = AC and h is the altitude from A to BC then

Lim 3 equals to [where  is the area and p the is perimeter of the triangle ABC]
h 0 p
(A) 1/128r (B) 1/215r
(C) 1/128 (D) 1/215
Sol.[C] AB = AC and AD = h
BC = 2BD = 2 OB2  OD 2

= 2 r 2  (h  r ) 2 = 2 2hr  h 2

AB = AD2  BD 2 = h 2  2hr  h 2

= 2hr
Perimeter p = 2AB + BC
= 2 2hr  h 2 + 2hr

Area of Triangle ABC  = 1/2 BC. AD

=h 2hr  h 2

 2r  h 1
Lim = Lim
h 0 p 3 h 0
8 2r  h  2r  3 =
128

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 n n

Q.1 Let  = nlim



r 1
r2
and  = nlim

 (r
r 1
3
 r2)
then
n3 n4
(A)  =  (B)  < 
(C) 4 – 3 = 0 (D) 3 – 4 = 0
n

Sol.[D]  = nlim

r
r 1
2

n3
n 2
r 1
 = nlim
 
r 1
  
 n  n
1
1  x3 
=  x 2 dx =  
 3  0
0

1
=
3
n n
r3 r2
=  r 1 n4
 
r 1 n4
0

[Sum of r given the expression in n3]

2

n n 3
r3 r 1
=  r 1 n4
=    
r 1 
n  n
1
1  x4 
=  x3 dx =  
 4  0
0

= 1/4

cos x
Q.1 If is a periodic function, then
sin ax
Lim Lim (1 + cos2m n!a) is equal to -
m n 

(A) 0 (B) 1 (C) 2 (D) – 1

cos x
Sol.[C] If is periodic then a must be rational
sin ax
A is rational n  

n!a will become and integral multiple of  Lim 2m

n  (1 + cos n!a)

1 + (± 1)2m  2

k k
k (1  x 2 )1 / 3  (1  2 x )1 / 4
Q.3 If r 1
cos–1 r =
2
for any k  1 and A =  r 1
(r)r. Then Lim
x A
x  x2
=

(A) 1/2 (B) 0 (C) A/2 (D) /2

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  
Sol.[A] For k = 1, cos–1 1 = 1 = cos =0
2 2
k = 2, cos–1 1 + cos–1 2 = 

 cos–1 2 = = 2 = 0
2
By induction j = 0, j = 1, …… k.
k
Hence A = r 1
(r)r = 0

(1  x 2 )1 / 3  (1  2 x )1 / 4
Required limit xlim
0
x  x2
2x 2
lim (1  x 2 ) 2 / 3  (1  2 x ) 3 / 4 1
x 0 3 4 =
2
1  2x

 2 2 
Q.6 Let f(x) = max.  x  , {sin x} . Which of the following is correct ?

 4 

(A) lim f(x) = 0 (B) xlim f(x) = 0

x 0  0

lim lim
(C)  f(x) = 0 (D)  f(x) = 1
x x
2 2
Sol.[D] By graph
y = x2 – 2/4

y = {sin x}

/2 /2

Q.5 The value of

lim cos (tan–1 (sin (tan–1x))) is equal to -
|x| 
(A) –1 (B) 2
1 1
(C)  (D)
2 2
Sol.[D] lim cos (tan–1 sin (tan–1x)
| x | 

 x 
lim cos (tan–1 sin sin–1  )
| x |   2 
 1 x 

 x 
lim cos tan–1  
| x |   2 
 1 x 

lim cos cos–1 1 x2

| x | 
2x 2  1

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 lim = 1 x2 1
| x |  =
2x 2  1 2

x
 e  1 x 
Q.1 Lim    is
x   1  e  e 1  x 
 1e   e  1 e
(A) e(1 – e) (B)   (C)   (D)
e e  e  1e  e e

Sol. [C]

1/ x
 f (1  x ) 
Q.1 Let f : R  R such that f(1) = 3 and f (1) = 6. Then Lim  equals –
x 0  f (1) 
(A) 1 (B) e1/2 (C) e2 (D) e3
Sol. [C]

Q.1 Lim n 2 ( n a  n 1 a ) ; a > 0, is

n 

(A) 1 (B) ea (C) n a (D) 0

Sol. [C]

log( x  a )
Q.3 Lim
x a  log(e x  e a )

(A) 1 (B) 0 (C) e (D) 1/e

Sol. [A]

1 x x 3
Q.7 The value of  so that xlim
0 2
(e – e – x ) = is
x 2
(A) 1 (B) 0 (C) 4 (D) 2
 e x  e x  1
Sol.[D] Using L Hospital Lim
x0
2x
Nr  0 for x  0
e0 – e0 – 1 = 0 or  = 2

Q.4 Lim n (1  x )  x  1 =
x 0 2
x x
1 1
(A)  (B) (C)  (D) 1
2 2

n (1  x )  x 2  x
Sol.[B] Lim
x 0 using expansion
x2
x2 x3
n(1 + x) = x –  ……...
2 3

x x
Q.5 Lim (cos  )  (sin  )  1 is equal to
x 2
x2
(A) 1
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 (B) cos n sin + sin2
(C) sin n cos + cos2
(D) None of these
Sol.[D] Using D. L.Hospital rule

x (n x ) 3
Q.6 Value of Lim
x 
is equal to
1 x  x 2
(A) 0 (B) 1
(C)– 1 (D) 3

(n x ) n
Sol.[A] xLim
  0 (m > 0)
xm

n2 n 3
n (1  kx ) ek x
1 lim f ( x ) is
Q.1 If f(n) = xlim
0 k 1
x
, g(n) = xlim
0 
k 1
x
, then x  
g( x )
(A) does not exist (B) 2
(C) 0 (D) data inadequate 
2
n 2 ( n 2  1)  n ( n  1) 
Sol.[B] f(n) = , g(n) =  
2  2 
x 2 ( x 2  1) 4
 xlim
 =2
2( x  1) 2 x 2

If P = nlim
2
Q.2 2 3 4 n–1 n 1 /( n 1) , then P4 equals
  (ea .e a …..e a )

(A) ea2 (B) e2a2

(C) ea (D) e2a
Sol.[C] P = nlim
2

(e1+3+…..n/2 terms a2+4+…..n/2 terms) 1 /( n 1)

= nlim
2

(en/4 (1+ n–1) an/4(2 + n)) 1 /( n 1)
n2 2n  n 2
= nlim
 4 ( n 2 1) 4 ( n 2 1) = e1/4 . a1/4  P4 = e.a
e a

1
Q.2 y= [(n + 1) (n + 2) …….2n]1/n
n
then Lim y is equal to
n 

4 2
(A) log (B) log 
e e
4 16
(C) (D) 
e e
Sol. [C]
1/ n
 n  1   n  2   2n  
y =   ........ 
 n   n   n 

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 1   n 1 n2 
log y = log   log   .....
n   n   n  
n
1  r
=
n  log1  n 
r 0
1
Lim log y =
n   n (1 + x)dx
0

Q.1 If 0 < a < b, then nlim


(bn + an)1/n is equal to
(A) e (B) a
(C) b (D) None of these
Sol. [C]

Q.3 x x  x  x  4 is
lim
x 1 x 1
1 15
(A) (B)
16 16
7 3
(C) (D)
8 4
Sol. [B]
x
 e   1 x 
Q.5 Lim     is
x   1  e   e 1  x 
 1e 
(A) e(1 – e) (B)  
e e 
 e  1 e
(C)   (D)
e  1e  e e

Sol. [C]

x
 1 1 1 1 
Q.3 Lim 1 x  2 x  3 x  ...  99 x  is
x 0   
 
(A) 1 (B) 99
(C) 99 × 50 (D) 0
Sol.[B]

 n n 1 n 2  1
Q.2 Lim1.
n  
 r 1

r  2.
r 1
r3 
r 1

r  .......  n.1 4 is
n

1 1
(A) (B)
12 6
1 1
(C) (D)
24 8
Sol.[C]

Q.2 If f(x) = nlim


n (x1/n–1), x > 0, then f(xy) is:
(A) f(x) f(y) (B) f(x) + f(y)

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 f (x)
(C) f(x) – f(y) (D)
f ( y)
x 1/ n  1
Sol.[B] f(x) = nlim

 f(x) = nx
1/ n

x nx
m  m
Q.7 Lim n C x   1   equals to -
n   n   n 

mx mx
(A) . e m (B) .em
x! x!

(C) e0 (D) 0
n
 m
n x 1  
m  n
Sol. [A] Lt 
n  x n  x n x  m x
1  
 n
x
n  m
=
mx 
Lt 1 
m
 n 1  n 
x n  n  Lt
n  nx nx

n x
m x m  m
= e Lt Lt 1  
x n  nx n  x n   n 

e m m x n
= Lt
x n   n  x nx

x 2  3x  5
Q.3 lim exists if
x  4x  1  x k
(A) k = 2 (B) k < 2
(C) k  2 (D) None of these
Sol.[C]
(1  3y  5 y 2 ) y k  2
Put x = 1/y  lim
y0 4 y k 1  y k  1
 limit will give a definite value if k  2
[x]
Q.6 The value of lim is (where [·] denotes greatest integer):
x 0 x
(A) 1 (B) 0
(C)  (D) None of these
[0  h ]
Sol.[B] lim 0
h 0 0  h
n x n x
x 1/ e
lim (2 )  (3 x )1/ e
Q.73 The value of x 
;
xn
(where n  N) is-
(A) 0 (B) n 2/3
(C) nn 3/2 (D) Not defined
Sol.[A]

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Q. 120 The figure shows a right-angled triangle with its hypotenuse OB along y-axis and its vertex along

y = x2. The value of xlim

0
(OB)

(A) 2 (B) 1
(C) 1/2 (D) 0
Sol.[B]
Q.9 C is a point on the circumference of a circle and D is the foot of the perpendicular from C on a fixed diameter AB.

CD 2
Then the limit of as C tends to B along the circumference -
DB
(A) does not exist
(B) equal to one
(C) is equal to the length AB
(D) None of these

lim CD 2
Sol.[C] pt . CB
DB
C

A B
D

lim AD.BD
pt . CB = AB
DB
1  f ( x )  f (3) 
Q.7 Let f (x) = , the value of Lim  is
18  x 2 x 3  x 3 
(A) 0 (B) –1/9 (C) +1/3 (D) 1/9
f ( x )  f (3)
Sol. [C] Lim Apply L-Hospital rule
x 3 x 3
f ( x )  0 1 1 1
Lim = f (3) = = =
x 3 1 18  9 9 3

lim 1  cos 2( x  1)
Q.1 x 1
x 1
(A) exists and it equal to 2

(B) exists and it equals – 2

(C) does not exist because x – 1  0

(D) does not exist because the left hand limit is not equal to the right hand limit

1  cos 2( x  1) 2 | sin( x  1) |
Sol.[D] xlim
1
= xlim
1
x 1 x 1

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 2 | sin( x  1) |
But xlim
1
x 1
= 2 x 1
lim sin( x  1) = 2 and
x 1
lim 2 | sin( x  1) |
x 1–
x 1
=– lim sin( x  1) = – 2 .
2 x 1–
x 1
Q.2 Let ƒ(x) = sgn (sgn (sgn x)). Then xlim
0
ƒ(x) is -

(A) 1 (B) 2
(C) 0 (D) None of these
Sol.[D] By definition we have for x  0, sgn (sgn x)
 x 
= sgn  
| x |
x/|x| x
= = = sgn x. Thus, sgn [sgn)
|x|/|x| |x|
 1 x0

(sgn x) = sgn x =  0 x  0
– 1 x  0

Therefore, xlim
0
ƒ(x) = 1 but xlim
0 –
ƒ(x) = –1.

 1 
Q.3 lim  4

1  3x  x 2 
3
x 4 1 
is -
x 1  x 2  x 1 1 x3  x 3  x 1 
  

(A) 3 (B) 2 (C) 4 (D) 28/3

 1 
 4 1  3x  x 2  x 4 1 
Sol.[A] xlim 
1 



3
2
 x  x
1
1 x3  x 3  x 1 


 1 
2  3x ( x 4  1) 
 4 x  1  3x  x
= xlim 
 x 1



1

3
1 x 3
 x 4  1 


 1 
2 
= xlim  4 x  1  3x  x   3x 
1  3
x 
1 
  

= xlim
1
[x – 1 + 3x] = 3.

lim 8  x2 x2 x2 x2 
Q.4 1  cos  cos  cos cos 
x 0
x 8  2 4 2 4 

is equal to -
1 1 1 1
(A) (B) – (C) (D) –
16 16 32 32

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 8  x2 x2 x2 x2 
Sol.[C] xlim
0
1  cos  cos  cos cos 
x 8  2 4 2 4 

8  x2 
  cos x
2 
1  cos x
2 
= xlim  1  cos 
0
x 8  2 
 4 
 2 


8  x2   2 
1  cos x 
= xlim 1  cos 
0
x 8  2 


 4 

8 2 2
2 x 2 x
= xlim
0
. 2 sin . 2 sin
x8 4 8
2 2
 2   x2 
 sin x  2  sin  2
32    x2    .  x 2 
= xlim 4   . 8
0
x8  x2   4   2   8 
 
 
 x 
 4   8 
1
= .
32
Hence (C) is the correct answer.
4 3
n5  2  n2 1
Q.5 The value of nlim

is -
5 2
n 4  2  n3 1

(A) 1 (B) 0 (C) –1 (D) 

4 3
n5  2  n2 1
Sol.[B] nlim
 5 2
n 4  2  n3 1

2 1
n5/ 4 4 1   n 2/3 3 1 
n5 n2
= nlim

2 1
n4/5 5 1   n 3/ 2 2 1 
4
n n3

n5/ 4 2 n 2/3 1
4 1  3 1
3/ 2 5 3/ 2
n n n n2
= nlim

n 4/5 2 n 3/ 2 1
5 1  2 1
n 3/ 2 n4 n 3/ 2 n3
[On dividing the numerator and denominator by the highest power of n i.e. n3/2]

1 1 21
4 1
 5/ 6 3 1 2
1/ 4 5 00
n n n n
= nlim

= = 0.
1 2 1 0 1
5 1  2 1 3
n 7 / 10 n4 n
Hence (B) is correct answer.
k 2
Q.6 lim n sin ( n!) , 0 < k < 1, is equal to -
n 
n2
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 (A)  (B) 1
(C) 0 (D) None of these

k n k sin 2 ( n!)
2
n sin ( n!)
Sol.[C] nlim

= nlim
 n  2
n2 1  
 n

sin 2 (n!)
= nlim
 n1 k  2
1  
 n
a finite quantity
=


[ sin2 (n!) always lies between 0 and 1. Also, since 1 – k > 0,  n1–k   as n  ]
= 0.
Hence (C) is the correct answer.
Q.7 In a circle of radius r, an isosceles triangle ABC is inscribed with AB = AC. If the ABC has perimeter p = 2 [
lim A is -
2hr  h 2 + 2hr ] and area A = h 2hr  h 2 , where h is the altitude from A to BC, then h 0
p3
1
(A) 128 r (B)
128 r

1
(C) (D) None of these
64 r

A h 2hr  h 2
Sol.[B] lim = lim 3
h 0 p 3 h 0 8
 2hr  h 2  2hr 

 

h. h 2r  h
= lim
h 0 8 h .h  2r  h  2r 3
2r  h
= lim
h 0 8  2r  h  2r 3
2r 1
= = .

8 2 2r  3
128r

Hence (B) is the correct answer.

Q.8 The limiting value of (cos x)1/sin x as x  0 is -
(A) 1 (B) e
(C) 0 (D) None of these
Sol.[A] Put cos x = 1 + y as ; x  0 ; y  0

Hence xlim
0
(cos x)1/sin x = ylim 1/y y/sin x
0 [(1 + y) ]

cos x 1  x
lim = lim   tan  = e0 = 1.
e x 0 sin x x 0  2
e
Hence (A) is the correct answer.

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 [ x ]  [ 2 x ]  ....  [nx ]
Q.9 If [.] denote the greatest integer function then nlim
 2
is -
n
2
(A) 0 (B) x (C) x/2 (D) x /2
Sol.[C] nx – 1 < [nx]  nx. Putting n = 1, 2, 3,…. , n and adding them,
x  n – n <  [nx]  x  n
n 1 [ nx ] n
x. 2 – < 2 x. 2 … (1)
n n n n

 n 1  n
Now, nlim
 
x.
2
  = x . nlim
 
 n n n2

1 x
– nlim

=
n 2
As the two limits are equal, by (1)

lim [ nx ] = x .
n 
n2 2
Hence (C) is correct answer.

{ƒ( x )}2 n  1
Q.10 Let ƒ(x) = x – [x], where [x] denotes the greatest integer  x and g(x) = nlim
 , then g(x) is equal
{ƒ( x )}2 n  1
to -
(A) 0 (B) 1
(C) –1 (D) None of these
Sol.[C] As 0  x – [x] < 1  x  R, 0  ƒ(x) < 1.

 lim 2n
n  {ƒ(x)} = 0.

{ƒ( x )}2 n  1
Thus, for x  R, g(x) = nlim

{ƒ( x )}2 n  1

0 1
= = –1.
0 1
Q.11 The value of
1/ 2
lim 2( x )  3( x )1 / 3  4( x )1 / 4  .....  ( x )1 / n is
x 
(2 x  3)1 / 2  (2 x  3)1 / 3  ....  (2 x  3)1 / n

1
(A) 2 (B) 2 (C) (D) 0
3
Sol.[A] Given

lim 2 x1 / 2  3x1 / 3  ....  nx1 / n

x  ( 2 x  3)1 / 2  (2 x  3)1 / 3  ...  ( 2 x  3)1 / n

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  1   1   1 
2 1 / 2   3 1 / 3   ....  n 1 / n 
= hlim h  h  h 
0 1 1 1
1/ 2
(2  3h )1 / 2  1 / 3 (2  3h )1 / 3  .....  1 / n (2  3h )1 / n
h h h
1
[On putting x = as x   , h  0]
h
1 1 1 1 1 1
        
2  3h  2 3
 4h  2 4
 ....  nh  2n
= hlim
0 1 1 1 1
     
1/ 2
( 2  3h )  h  2 3  ( 2  3h )1 / 3  ....  h  2 n  ( 2  3h )1 / n

2  0  0  0  .......
= 1/ 2 = 2 .
2  0  0  .......
Hence (A) is the correct answer.
k k
k (1  x 2 )1 / 3  (1  2 x )1 / 4
Q.12 If  cos 1 r = for any k  1 and A =  ( r ) , Then xlim
r
 A
=
r 1 2 r 1 x  x2
(A) 1/2 (B) 0 (C) A/2 (D) /2
 
Sol.[A] For k = 1, cos –1 1 =  1 = cos =0
2 2

For k = 2, cos–1 1 + cos–12 =   cos–1 2 =  2 = 0
2
So by induction it can be shown that j = 0, j = 1……..k.
k
Hence A =  ( r ) r = 0.
r 1

Thus required limit

(1  x 2 )1 / 3  (1  2x )1 / 4
= lim
x 0 x  x2
2x 2
(1  x 2 ) 2 / 3  (1  2x ) 3 / 4 1
= = .
lim 3 4
2
x 0 1  2x
1  9 
Q.13 If f(n + 1) = f (n )  , n  N and
2  f (n ) 

f(n) > 0 for all n  N then nlim

 f(n) is equal to-

(A) 3 (B) –3
(C) 1/2 (D) None of these

Sol.[A] As n   . nlim lim

 f(n) = n  f(n + 1) = k say

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 1  9  1  9 
We have f(n + 1) =  f ( n )   or lim f(n) = lim f(n + 1) =  f (n )  
2  f (n )  n   n   2  f ( n ) 

1  9
k=  k    k2 = 9 or k = 3
2  k

 nlim
 f (n) = 3.

cos 2 (1  cos 2 (1  cos 2 (.........  cos 2 ( x ))).......)

Q.14 xlim 
  x  4  2    is equal to -
0 sin  
 
 x 


 4  2
(A) (B) (C) (D)
4  2 
Sol.[B] Let P

cos 2 (1  cos 2 (1  cos 2 (......... cos 2 ( x ))).......

= xlim 
  x  4  2  
0 sin  
  x 
  

cos 2 sin 2 (1  cos 2 (......... cos 2 ( x )).....

= xlim 
  x  4  2  
0 sin  
  x 
  

cos 2 (sin 2 (sin 2 (......... (sin 2 ( x )))).......


  x  4  2  
sin  
  
lim
= x 0  
 x    ( x  4)  2  
  
 ( x  4)  2    x  
   
 x 
 

cos 2 (sin 2 (sin 2 ).........(sin 2 ( x )).......) ( x  4)  2

·

  x  4  2   
sin 
 
= xlim
0

  x 
 ( x  4)  2 
 
 x 
 

cos 2 0 ( 2  2) 4
= . = .
1  
Q.15 Consider a series of number defined as following
x0 = a , x1 = a a, x2

= a a a , ……. Where a > 0 then nlim x =

 n

1  4a  1
(A) a (B)
2

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 1  4a  1
(C) (D) can not be found
2a

Sol.[B] We have x 2
n = a + xn–1

It is easy to see that the variable x n increases. Let us show that all its values remain less than some

constant number we have x 2

n –1 – x n–1 – a < 0 ( xn–1 < xn)

 (4a  1)  1   1 4a  1 
Hence,  x n 1  x
  n 1
–
 <0
 2   2 

( 4a  1)  1
Since the expression in the second bracket is positive, so we have xn–1 <
2

Put nlim x = nlim

 n–1
x = 
 n

From the original relation between xn and xn–1, we get

1  1  4a (4a  1)  1
2 –  – a = 0,  = and since   0 we have  = .
2 2

1
Q.1 lim
Let ƒ(x) = n   3 2n
. Then the set of values of x for which f(x) = 0, is -
1 
 tan 2 x   5
 
(A) |2x| > 3 (B) |(2x)| < 3

(C) |2x|  3 (D) |2x|  3

2
3 
Sol.[A] ƒ(x) = 0 if and only if,  tan 1 2 x  1
 

 
 tan–1 2x > or tan–1 2x < –
3 3

 2x > 3 or 2x < – 3

 |2x| > 3

Hence (A) is the correct answer.

Q.2 If I1 = xlim tan 1 x sin 1 x and I = lim sin 1 x tan 1 x , where | x | < 1, then which of the
0  2
x 0 
x x x x
following statement is true -
(A) Neither I1 nor I2 exists
(B) I1 exists and I2 does not exists
(C) I1 does not exists and I2 exists
(D) None of these

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 tan 1 x sin 1 x
Sol.[C] We know < 1 and > 1,  x  R
x x

tan 1 x sin 1 x
 – <0
x x

sin 1 x tan 1 x
and – >0
x x
 I1 does not exists and I2 exists
Hence (C) is the correct answer.
[ x ]  [ 2 x ]  ....  [nx ]
Q.3 If [.] denote the greatest integer function then nlim
 2
is -
n
2
(A) 0 (B) x (C) x/2 (D) x /2
Sol.[C] nx – 1 < [nx]  nx. Putting n = 1, 2, 3,…. , n and adding them,
x  n – n <  [nx]  x  n
n 1 [ nx ] n
x. 2 – < 2 x. 2 … (1)
n n n n

 n 1 
Now, nlim
 
x.
2
 
 n n

n 1 x
= x . nlim
 2
– nlim

=
n n 2
As the two limits are equal, by (1)

lim [ nx ] = x .
n 
n2 2
Hence (C) is correct answer.
1
lim x 4 sin  x 2
Q.1 Evaluate x –  x
1 | x |3
(A) –1 (B) 1
(C) 0 (D) None of these
Sol.[A] Let x = –y. Then y   when x  –
–1
lim y 4 sin  y2
 limit = y y
1 | – y |3
1
lim  y 4 sin  y 2
= y
y
1  y3
[ y is positive]

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  1 
 sin 3 2 
 y y y 
= ylim
  1
·  
3
 1 y 1  y3 

 y 

 
 sin 1 
 y  1 1 
= ylim
  1
·  
 1  
 1 y 1  1 
 y y3  y3 
  
1
=1. + 0 = –1.
1
Q.2 If ƒ(1) = g(1) = 2 and ƒ(1), g(1) exist then evaluate

lim ƒ(1)g ( x )  ƒ(1)  g (1)ƒ( x )  g (1)

x 1 g ( x )  ƒ( x )
(A) 1 (B) 2
(C) 3 (D) None of these
0
Sol.[B] Here, the form is .
0
ƒ(1)g( x )  g (1)ƒ( x )
 limit = xlim
1 
g( x )  ƒ( x )
{using L’ Hospital’s rule}
2{g( x )  ƒ( x )}
 xlim
1 = xlim
1 2 = 2.
g( x )  ƒ( x )
sin x
Q.3 Evaluate : xlim  sin x  x sin x .
0  
 x 
–1
(A) e (B) e
(C) e2 (D) None of these
sin x
x
Sol.[A] Limit = xlim
0  sin x  1 sin x
  x
x  
sin x
Put = t; then t  1 when x  0.
x
t
 limit = lim , which is of the form 1
t 1 t 1 t
1
t
= lim log t 1 t = lim log t
e t 1 e t 1 1 t

1 log t
= lim
e t 1 1

(Using L’ Hospital’s rule)

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 = e–1.
x
 x 2  5x  3 
Q.7 If ƒ(x) =   then xlim

ƒ(x) is -
 x2  x  2 
 
(A) e–4 (B) e3 (C) e2 (D) e4
x
 x  5x  3 
2
Sol.[D] xlim

ƒ(x) = xlim 
  2


 x x2 
x
 4x  1 
= xlim
 1  2 
 x x2

 x2 x 2 
= xlim
 4 x  1  4 x 1 

1 
 
  … (1)
 x2  x  2  

 


x (4 x  1) 4 1/ x
Where  = =  4 as x  
2
x x2 11/ x  2 / x 2

(1)  xlim 4
 ƒ(x) = e .

sin x , x  n 
x 2  1, x2
Q.15 Let ƒ(x) =  , where n  I and g (x) =  , then xlim
0
g [ƒ(x)] is -
 2, x  n 
 3, x2

(A) 1 (B) 0
(C) 3 (D) Does not exist

[f ( x )]2  1, x2
Sol.[A] g [ƒ(x)] = 

 3, x2

 g [ƒ (x)] = sin2 x + 1, x  n
3, x = n

R.H.L. = hlim
0
g [ƒ(0 + h)] = hlim
0
(sin2 h + 1) = 1

L.H.L = hlim
0
g [ƒ(0 – h)] = hlim
0
(sin2 h + 1) = 1

 hlim
0 g[ƒ (x)] = 1.

Q.11 Given a real valued function ‘f’ such that

 tan 2 {x}
 for x  0
 x 2  [ x ]2

f(x) =  1 for x  0
 {x} cot{x} for x  0



2
 Lim f ( x ) 
then the value of cot–1   is –
 x 0 

(A) 0 (B) 1 (C) –1 (D) None

Sol.[D] Lim– f(x) = Lim

h 0
{ h} cot{ h}
x 0
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 = Lim
h 0
(1  h ) cot(1  h ) = cot 1

2 2
Lim f(x) = Lim tan {h} = Lim tan h = 1
 h 0 h 0
x 0 2 2 2
h  [h ] h

 Lim
h 0 f(x) does not exist.

f ( x  c) f ( x  2c) f ( x  3c)
Q.12 Let g(x) = f ( c) f (2c) f (3c)
f (c) f (2c) f (3c)

g( x )
where c is constant then Lim
x 0
is equal to
x
(A) 0 (B) 1 (C) –1 (D) f(c)
Sol.[A] We note that g(0) = 0
g( x ) 0
 Lim
x 0 is form
x 0

g( x ) g( x )
 Lim
x 0 = Lim
x 0 = g(0) = 0.
x 1
Q.14 If  and  are the root of the quadratic equation ax2 + bx + c = 0,

lim 1  cos(cx 2  bx  a )
then x  1 =
 2(1  x ) 2

c 1 1 c 1 1
(A)    (B)   
2 
   
 2 
   


c 1 1
(C)    (D) None of these
 
   


a b 1 1
Sol.[A]  ax2 + bx + c = 0 has roots  and  then   c  0 i.e., cx2 + bx + a = 0 has roots and .
x 2 x  

2 bx a   1  1
 cx    = c x    x  
 c c    

lim  1  cos(cx 2  bx  a ) 
= x 1  
  2 (1   x ) 2 
 
  2 
 sin 2  cx  bx  a 
lim   2 
 
= x 1  
  (1  x ) 2 
 

 


 cx 2  bx  a 
sin  
lim 
 2 

= x 1 (1  x )


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 c  1  1 
sin   x    x  
lim 2     
= x 1
  1
  x  
  

c 1  1  c 1
sin   x    x     x  
2
      2 
= lim · lim
x
1 c 1  1  x
1  
  x    x   
2   

c  1 1

  

2   c 1 1
= 1· =    .
 2    



Lim 1  cos 3 x
Q.89 x 0
equals
x. sin x. cos x
1
(A) (B) 1
4
3
(C) (D) Does not exist
2
(1  cos x ) x (1  cos x  cos 2 x )
Sol.[C] Lim
x 0  
x2 sin x cos x
1 111 3
=  1 
2 1 2

x
Q.90 Lim a  1 equals
x 0
x
(A) log a (B) 1
(C) 0 (D) Does not exist
Sol.[A] By D.L. Hospital rule
x
Lim a log a = a0. log a = log a
x 0
1

Q.91 Lim ( 2 x  3)( x  1) equals

x 1
2x 2  x  3
1 3
(A) (B)
2 2
(C) 1 (D) – 1

( 2 x  3)( x  1) x 1 1
Sol.[A] Lim
x 1
 Lim 
( x  1)(2 x  3) x 1 ( x  1)( x  1) 2

Lim  1 8 n3 
Q.92   .....  equals
n  4 4 4
1  n 1 n 1  n 

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 1 1
(A) (B)
4 8
1
(C) (D) None of these
2
2 2
13  2 3  .....  n 3 Lim n ( n  1)
Sol.[D] Lim
n 
= n 
1 n4 4(1  n 4 )
2
 1
n 4 1  
= Lim  n  = (1  0) 2  1
n 

 1  4(0  1) 4
4 n 4  4  1
n 

Q.93 Lim (6n + 5n)1/n equals

n 

(A) 6 (B) 5
(C) 5/6 (D) e
1/ n 0
 n   5 

Sol.[A] 6 Lim 1   5    61    
n  
  6   
  6  

= 6 (1 + 0)0
= 6 × 10
= 6 × 1 = 6.

Q.94 Lim cos 2  cos 2 x equals

x  –1
x2  | x |
(A) sin 2 (B) 2 sin 2
(C) 4 sin 2 (D) 0

Sol.[B] Lim cos 2  cos 2 x

x  1
x2  x
2 sin(1  x ) sin( x  1)
= xLim
 1 x ( x  1)

Lim 2 sin(1  x ) . sin( x  1)  2 sin( 2) × 1

x  1 (1  x ) x (1)
= 2 sin 2
Q.99 If  is the distance between orthocentre & circumcentre of triangle with vertices (1, 0),

 1   
  , 3 ,   1 ,  3  then Lim [x] is
 2 2   2 2  x 
   
(A) 0 (B) – 1
(C) 1 (D) Does not exist
Sol.[D] If A,B,C are vertices then AB = BC = CA= 3 ABC is equilateral
 Required distance = 0 = 

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  Lim
x  0 [x] = Does not exist
3
Q.100 Lim x (log x ) equals
x 
1 x  x2
(A) 0 (B) – 1
(C) 1 (D) Does not exist
1
Lim (log x ) 3  x.3(log x ) 2 
Sol.[A] x  x
1  2x
(By D.L. Hospital rule)
1 1
Lim 3(log x ) 2   6(log x ) 
 x  x x
2
(By D.L. Hospital rule)
3(log x ) 2  6 log x
 Lim
x 
2x
(By D.L. Hospital rule)
1 6
6 log x  
 Lim
x  x x
2
(By D.L. Hospital rule)
6 log x  6
 Lim
x  2x
(By D.L. Hospital rule)
1
6   0
 Lim
x  x
2
(By D.L. Hospital rule)
6
=0
 
2

lim 1
Q.74 x a ( n N) equals -
( x – a ) 2 n –1
(A)  (B) – 
(C) 0 (D) Does not exist
Sol.[D] Do your self.

2x 3 – 4x  7
Q.75 The value of xlim
 is-
3 2
3x  5 x – 4
(A) 2/3 (B) –7/4
(C) –4/5 (D) 
Sol.[A] Do your self.
 x 
lim
Q.76 x   3  equals-
 x 3  10 

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 (A) 1 (B) 2
(C) 3 (D) None of these
Sol.[A] Do your self.
Q.77 lim n[a1/n–1] equals-
n 
(A) a (B) logea
(C) 1 (D) None of these
Sol.[B] Do your self.
lim x
Q.78 x  equals -
x x x
(A) 0 (B) 1
(C)  (D) None of these
Sol.[B] Do your self.
log (3  x )  log (3  x )
Q.79 If xlim
0
= k, the value of k is -
x
2
(A) – (B) 0
3
1 2
(C) – (D)
3 3
Sol.[D] Do your self.
Q.80 If f(x) is a differentiable function and

f (2) = 6, f (1) = 4, f (c) represents the differentiation of f(x) at x = c, then

2
lim f (2  2h  h )  f ( 2)
h 0
f (1  h 2  h )  f (1)

(A) may exist (B) will not exist

(C) is equal to 3 (D) is equal to –3

Sol.[C] Do your self.

     
Q.111 xlim

x cos   sin   is -
 4 x   4x 
 
(A) (B)
2 4
(C) 1 (D) None of these [B]
lim 1  cot x 3
Q.112  is -
x
4 2  cot x  cot 3 x
11 3
(A) (B)
4 4
1
(C) (D) None of these [B]
2

Q.113 lim 1  x 4  (1  x 2 ) is –
x 
x2
(A) 0 (B) – 1 (C) 2 (D) – 2 [A]

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Q.114 lim  3x 
 9x 2  x 
 equals?
x    

1 1 1 1
(A) (B) (C) – (D) – [B]
3 6 6 3

lim n!
Q.115 is equal to -
n  ( n  1) !  n !
(A) 0 (B) 
(C) 1 (D) None of these [A]

Q.116 The value of xlim


x1/x is -

(A) 0 (B) 1
(C)  (D) None of these [B]
 x3 1 
Q.121 If xlim 
  x 2  1
 ( ax  b )  = 2, then -

 
(A) a = 1, b = 1 (B) a = 1, b = 2
(C) a = 1, b = – 2 (D) None of these [C]
log( x  a )
Q.122 The value of lim is -
x a log(e x  e a )
(A) 1 (B) – 1
(C) 0 (D) None of these [A]
3
lim x 2 1  x3 1
Q.123 x  4
equals -
5
x4 1  x4 1
(A) 1 (B) 0
(C) – 1 (D) None of these [B]
2
x
Q.124 lim e  cos x is -
x 0
x2
3 1
(A) (B)
2 2
2
(C) (D) None of these [A]
3
sin px
Q.125 If xlim
0 tan 3x
= 4, then p is equal to -

(A) 6 (B) 9 (C) 12 (D) 4 [C]

 x 3 
Q.126 The value of xlim  log a
3 
 is -
 x  6  3 
(A) loga 6 (B) loga 3
(C) loga 2 (D) None of these [A]

Q.127 lim ([x – 3] + [3 – x] –x), where [.] denotes the greatest integer function, is equal to -
x 3

(A) 4 (B) – 4
(C) 0 (D) None of these [B]

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  4 
Q.128 The value of xlim

3x sin  x  is -
3 
(A) 4 log 3 (B) 3 log 4
(C) 4 (D) None of these [C]

Q.129 If xlim
0
(1 + ax)b/x = e2, (a, b  N), then -

(A) a = 4, b = 2 (B) a = 8, b = 4
(C) a = 16, b = 8 (D) None of these [D]
 sin(1  [ x ])
 , [x]  0
Q.130 If f(x) =  [x] then lim f(x) is -
x 0

 0, [x]  0
(A) – 1 (B) 0
(C) 1 (D) None of these [B]

Q.131 lim log (1 / 4 1 / 8 1 / 16 ......to n terms)

is equal to -
n  0.2 5

(A) 2 (B) 4 (C) 8 (D) 0 [B]

Q.132 If f(x) = [2x], then -

(A) lim f(x) = 0 (B) lim f(x) = 1

x 1/ 2  x 3 / 4 

(C) xlim
1 / 2
f(x) = 0 (D) x lim
3 / 4
f(x) = 2 [B]

Q.141 The value of

 3 1
 x  2x 
1 
 x  4 x  2  is -
lim    
x  2  x 3  8   x2 x 2  
   
(A) 1/2 (B) 2
(C) 1 (D) None of these [A]

Q.142 lim [log n–1(n).log n(n + 1).log n+ 1(n + 2)… log k (nk)] is equal to -
n  n 1

(A)  (B) n
(C) k (D) None of these [C]
n
12  2 2  32  ...  r 2
Q.143 If tr =
3 3 3
1  2  3  ....  r 3
and Sn = r 1
(–1)r.tr, then nlim S is given by -
 n

2 2 1 1
(A) (B) – (C) (D) – [B]
3 3 3 3
x x  x 
Q.144 The value of nlim

cos   cos   …cos  n  is-
2
  4
  2 
sin x
(A) 1 (B)
x
x
(C) (D) None of these [B]
sin x

Q.145 lim (1+ x) (1 + x2) (1 + x4) ….. (1 + x2n), | x | < 1, is equal to -

n 

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 1 1
(A) (B)
x 1 1 x
(C) 1 – x (D) x – 1 [B]
2 2 2 2
Q.146 lim tan[e ]x  tan[ e ]x is,
x 0
sin 2 x
(where [.] is G.I.F.) -
(A) 0 (B) 8
(C) 15 (D) None of these [C]

Q.78 lim x cos    sin    is -

x 
 4x   4x 
 
(A) (B)
2 4
(C) 1 (D) None of these [B]
lim 1  cot x 3
Q.79  is -
x
4 2  cot x  cot 3 x
11 3
(A) (B)
4 4
1
(C) (D) None of these [B]
2

Q.80 lim 1  x 4  (1  x 2 ) is –
x 
x2
(A) 0 (B) – 1 (C) 2 (D) – 2 [A]

Q.81 lim  3x 
 9x 2  x 
 equals?
x    

1 1 1 1
(A) (B) (C) – (D) – [B]
3 6 6 3

lim n!
Q.82 is equal to -
n  ( n  1) !  n !
(A) 0 (B) 
(C) 1 (D) None of these [A]

Q.83 The value of xlim


x1/x is -

(A) 0 (B) 1
(C)  (D) None of these [B]

 x 2  k, when x  0
Q.84 Let f(x) =  2 . If the function f(x) be continuous at x = 0, then k =

 x  k , when x  0
(A) 0 (B) 1
(C) 2 (D) – 2 [A]
 x 1 3 
Q.88 If xlim


 2
 (ax  b)  = 2, then -
 x 1 

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 (A) a = 1, b = 1 (B) a = 1, b = 2
(C) a = 1, b = – 2 (D) None of these [C]
log( x  a )
Q.89 The value of lim is -
x a log(e x  e a )
(A) 1 (B) – 1
(C) 0 (D) None of these [A]
3
lim x 2 1  x3 1
Q.90 x  4
equals -
5
x4 1  x4 1
(A) 1 (B) 0
(C) – 1 (D) None of these [B]
2
x
Q.91 lim e  cos x is -
x 0
x2
3 1
(A) (B)
2 2
2
(C) (D) None of these [A]
3
sin px
Q.92 If xlim
0 tan 3x
= 4, then p is equal to -

(A) 6 (B) 9 (C) 12 (D) 4 [C]

 x 3 
Q.93 The value of xlim  log a
3 
 is -
 x  6  3 
(A) loga 6 (B) loga 3
(C) loga 2 (D) None of these [A]

Q.94 lim ([x – 3] + [3 – x] –x), where [.] denotes the greatest integer function, is equal to -
x 3

(A) 4 (B) – 4
(C) 0 (D) None of these [B]
 4 
Q.95 The value of xlim

3x sin  x  is -
3 
(A) 4 log 3 (B) 3 log 4
(C) 4 (D) None of these [C]

Q.96 If xlim
0
(1 + ax)b/x = e2, (a, b  N), then -

(A) a = 4, b = 2 (B) a = 8, b = 4
(C) a = 16, b = 8 (D) None of these [D]
 sin(1  [ x ])
 , [x]  0
Q.97 If f(x) =  [x] then lim f(x) is -
x 0

 0, [x]  0
(A) – 1 (B) 0
(C) 1 (D) None of these [B]

Q.98 lim log (1 / 4 1 / 8 1 / 16 ......to n terms)

is equal to -
n  0.2 5

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 (A) 2 (B) 4 (C) 8 (D) 0 [B]
Q. 99 If f(x) = [2x], then -

(A) lim f(x) = 0 (B) lim f(x) = 1

x 1/ 2  x 3 / 4 

(C) xlim
1 / 2
f(x) = 0 (D) x lim
3 / 4
f(x) = 2 [B]
x
Q.75 lim  x 2  5x  3  =
x   x x3 
2
 
(A) e4 (B) e2
(C) e3 (D) e [A]
n
sin x
Q.105 For m, n  I+, Lt is equal to -
x 0 (sin x ) m
(A) 1, if n = m (B) – 1, if n > m
n
(C) (D) None of these [A]
m
2f ( x )  3f (2 x )  f ( 4 x )
Q.109 Let f ''(x) be continuous at x = 0 and f '' (0) = 4. Then value of lim is -
x 0 x2
(A) 12 (B) 10
(C) 6 (D) 4 [A]

Q.110 The value of lim 1 2  x  3 is -

x 2 x2
1 1
(A) (B)
8 3 4 3
(C) 0 (D) None of these [A]
x  sin x
Q.111 lim is equal to -
x 0 x  sin 2 x
(A) 1 (B) 0
(C)  (D) None of these [B]
1  sin x  cos x  log (1  x )
Q.117 lim equals -
x 0 x3
(A) 1/2 (B) – 1/2 (C) 0 (D) 1 [B]

(1  cos 2 x ) sin 5x
Q.118 lim =
x 0 x 2 sin 3x
(A) 10/3 (B) 3/10 (C) 6/5 (D) 5/6 [A]
20

Q.120 x
k 1
k
 20
=
lim
x 1 x 1
(A) 20 (B) 210
(C) does not exist (D) None of these [B]
Q.128 If x1, x2 are real and distinct roots of
ax2 + bx + c = 0, then

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 lim (1  sin (ax 2  bx  c))1 / x  x1 is equal to -
xx 1

(A) e x1  x 2 (B) e x 2  x 1

(C) e a ( x 1  x 2 ) (D) e a ( x 2  x1 ) [C]

log e x n  [ x ]
Q.129 lim , n  N, ([x] denotes greatest integer less than or equal to x) is equal to -
x  [x]
(A) – 1 (B) 0
(C) 1 (D) does not exist [A]
tan x sin x
a a
Q.130 lim (a > 0) is equal to -
x  0 tan x  sin x

(A) – lna (B) lna

(C) 1 (D) – 1 [B]
1/ x
x (3e  4)
Q.131 lim =
1/ x
x 0 2e
(A)  (B) 0
(C) does not exist (D) None of these [B]
2
x sin (1 / x )  x
Q.133 The value of lim is -
x  1 | x |
(A) 0 (B) 1
(C) – 1 (D) does not exist [A]
 sin x 
Q.134 lim , ( []  G.I.F.), is equal to -
x  
 x 
(A) 1 (B) 0
(C) does not exist (D) None of these [B]
 sin x  x  1
Q.136 lim   sin
x 0  x  x
(A) is equal to 1
(B) does not exist
(C) is equal to 0
(D) exists and different from 0 and 1 [C]
sin [cos x ]
Q.137 lim ([]  denotes greatest integer function) equals -
x  0 1  [cos x ]

(A) 1 (B) 0
(C) does not exist (D) None of these [B]

Q.138 lim (sec x  tan x ) is equal to -

x / 2

(A) 2 (B) – 1 (C) 1 (D) 0 [D]

Q.139 If xlim  ( x ) = a3, a  0, then lim   x  is equal to -

0 x 0 a
1 1
(A) (B) a3 (C) a2 (D) [B]
a3 a2

Q.140 The value of

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 x n  x n 1  x n  2  ...  x 2  x  n
lim is -
x 1 x 1
n ( n  1)
(A) (B) 0 (C) 1 (D) n [A]
2
sin (  cos 2 x )
Q.144 lim =
x 0 x2
(A)  (B) 
(C) –  (D) does not exist [C]

Q.145 If f(x) is a polynomial satisfying f(x) f(1/x) = f(x) + f(1/x) and f(2) > 1 then xlim
1
f ( x ) is -

(A) 2 (B) 1
(C) – 1 (D) None of these [A]
Q.146 Let f(x) = [x] + [– x]. Then for any integer m and non integer a, consider the following statements
(i) xlim
m
f ( x ) exist

(ii) f(x) is continuous at x = m

(iii) f(x) is continuous at x = a
(iv) xlim
a
f ( x ) exists.

Then correct statements is/are -

(A) (i), (ii), (iii) (B) (i), (ii)
(C) (i), (iii), (iv) (D) all of these [C]

cos x x 1
f (x)
Q.147 If f(x) = 2 sin x x2 2 x , lim is equal to -
x 0 x
tan x x 1
(A) 1 (B) – 1 (C) 0 (D) 2 [C]
 1 
Q.148 lim  2  cot x  =
x 0 x 
(A) 1 (B) 0
(C)  (D) does not exist [C]

 sin 2   sin 2  
Q.149 lim  =
  
  2  2 
(A) 0 (B) 1
sin  sin 2
(C) (D) [D]
 2

1 
Q.150  2 (1  cos 2 x ) is
 
lim
x 0 x
(A) 1 (B) –1
(C) does not exist (D) None of these [C]

f ( x )e nx  g ( x )
Q.151 If x > 0 and g is a bounded function, then lim is
n  e nx  1

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 (A) 0 (B) f(x)
(C) g(x) (D) None [B]
x.2 x  x
Q.152 lim equals
x  0 1  cos x

1
(A) log 2 (B) log 2
2
(C) 2 log 2 (D) None [C]
n
Q.153 lim log x  [ x ] , n  N, ([x] denotes greatest integer less than or equal to x)
x 0 [x]
(A) has value – 1 (B) has value 0
(C) has value 1 (D) does not exist [D]
Lim
Q.154  (x tan x –(/2) sec x) equals to
x
2
(A) –1 (B) 0 (C) 1 (D) 2
Sol. [A] xLim
 / 2
x. tan x – /2 sec x From – 
x. sin x –  / 2
 xLim
 / 2 . cos x
0  sin(  / 2)
Form 0/0  = –1
– sin(  / 2)
x – 2a  x – 2a
Q.155 The value of limit
x 2 a 2 2
is
x – 4a
1 1
(A) (B)
a 2 a
1 1
(C) (D)
3 a 4 a
1 / 2 x – 2a  1 / 2 x
Sol. [B] Apply L'H.Rule  xlim
2 a 2 2
1 / 2 x – 4a  2 x
1 ( x  x – 2a ) 1 4a 1
= lim . x 2 – 4a 2 = . =
2 x 2 a x – 2a x .x 2 2a 2 a
 g( x ) – 1
 ; x 1
Q.156 If f(x) =  x –1 and g (1) = 2, g(1) = 1. Then lim
x 1
f(x) is :
1 ; x 1

(A) 1 (B) 3(C) 2 (D) 4
1 1
Sol.[C] = lim g( x ) / x = lim g( x ) x = 2
x 1 2 g ( x ) 2 x 1 g( x )

( 2 x  1) 40 ( 4 x – 1) 5
Q.157 The limit xlim
 45
is
(2 x  3)
(A) 16 (B) 32 (C) 24 (D) 8
40 5
( 2 x  1) ( 4 x – 1) 2 .45 45
40
Sol.[B] xlim
 45
= = = 25 = 32
( 2 x  3) 2 45
25

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 lim sin x  (sin x ) sin x
Q.158 x  / 2
is equal to-
1  sin x  n sin x
(A) 4 (B) 2
(C) 1 (D) None of these
Sol. [B] take sin x = t
1  t t (1  nt )
lim t  (t) t
t 1
By L' Hospital = lim
t 1 1
1  t  nt 1 
t
t t 2
lim  {t (1 / t )  t (1  nt ) } = 2
t 1
1/ t 2

2x 3x
Q.159 lim 3  2 is equal to-
x 0
x
9 8
(A) log (B) log
8 9
9
(C) –log (D) log 72
8

 3  1 2  1
2x 3x
 lim
Sol.[A] xlim
0 
  = x 0

 x x 
 2x
3 1 2 3x  1 

 2  3

 2 x 3x 

9
= 2 log 3 – 3 log 2 = log 9 – log 8 = log  
8
Q.160 lim log 2 (tan2 2x) is equal to-
x 0 tan x

(A) 2 (B) e2
(C) 1 (D) e
log (tan 2 2 x )   
Sol. [C] xlim
0    By L' hospital
 
log tan 2 x
Q.161 If A and B are square Matrices of order 3 such that |A| = –1 , |B| = 3 then |3AB| = ---------
(A) –9 (B) –81 (C) –27 (D) 81
Sol. [B] |3AB| = 3 × 3 × 3 |A| |B| = 27 (–1) (3) = –81
4 2
Q.162 If A =   , then (A –2I) (A –3I) =
 1 1 
(A) A (B) I (C) 0 (D) 5I
  4 2 1 0     4 2  1 0 
Sol. [C]     2   
    3  
  1 1  0 1      1 1  0 1  
 2 2 1 2  0 0 
=    =   =0
  1  1   1  2 0 0
  
Q.163 If   is square root of I2, then ,  and  will satisfy the relation-
   
(A) 1 + 2 +  = 0 (B) 1 –2 +  = 0
(C) 1 + 2 –  = 0 (D) 2 +  = 1
2
   1 0       1 0
Sol. [D]  =   =
   

0 1
           0 1

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  2   0  1 0
  =  2 +  = 1
 0
2
    0 1

x 2 y2
Q.164 A circle of radius r is concentric with an ellipse  = 1. If common tangent is inclined to the major axis at
a 2 b2
an angle of , then tan2 equals-
r2  b2 r 2  b2
(A) (B)
a 2  b2 a2  r2
r 2  b2 r2  a2
(C) (D)
r2  a2 b2  r 2

Sol. [B]

equation of ellipse
x 2 y2
 = 1 Equation of circle x2 + y2 = r2
a 2 b2
Equation of tangent to ellipsey = mx ± a 2m2  b2
Equation of tangent to circle y = mx ± r 1  m2
for common tangent ± r 1  m 2 = ± a 2 m 2  b 2
r2 + r2m2 = a2 m2 + b2  a2m2 – r2m2 = r2 – b2
r 2 – b2 r 2 – b2
m2 =  tan 2
 =
a2 – r2 a2 – r2
lim x2 1  3 x3 1
Q.165 x  4
is equal to-
x 4 1  5 x 4 1
(A) 1 (B) –1
(C) 0 (D) 2
Sol. [C]
Q.166 lim { lim (92m ( n x))} where x  R then-
m  n 
(A) 1 if x  Rational (B) 0 if x is irrational
(C) 1 if x is irrational (D) 0 if x is rational.
Sol. [ ]
e sin x  1
Q.167 Find value of xlim
0
x

 (sin x ) 2 
lim 1  sin x   ....  1
Sol. x 0  2 

x
 
lim sin x 1  sin x  .....
x 0 2
x  

 
lim sin x . lim 1  sin x  ..... = 1.1 = 1
x 0 x 0  2 
x  

Q.168 lim x ( x  c – x )
x 

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Sol. lim x ( xc – x )
 xc  x 
x 
xc  x
lim x  x  c  x c x c
 xlim
x 
xc  x  x  1 c / x 1  =
2
x (1  a cos x )  b sin x
Q.169 If xlim
0 3
= 1 then find a, b
x
  x2   x3 
 
Sol. lim x 1  a 1  2  ....   b  x  3  .... = 1
x 0     
x3
3
lim x (1  a  b)  x (b / 6  a / 2).... = 1
x 0
x3
This is only possible when
1+a–b=0 ....(1)
b/6 – a/2 = 1 ....(2)
From (1), (2) a = –5/2, b = –3/2

Q.170 lim [sin x ] , where [.] is GIF, is

x0

(A) 0 (B) –1
(C) 1 (D) does not exist
LHL  1 
Sol.[D] As  does not exist
RHL  0 

 x 100  x 99  ......  4 
Q.171 The limit, lim   is

x 
 x 100  x 
(A) 1 (B) 10
(C) 100 (D) None of these

1 0 
Sol.[A] As the degree of Num. & den. is same so take common x100 we get lim   1
x  1  0 

1 1 
1 1 
Q.172 lim  + + +....... .........+  is equal to -
n   n n2  n n 2  2n n2  ( n – 1) n 


(A) 2 + 2 2 (B) 2 2 – 2
(C) 2 2 (D) 2 [B]

( x  1)10  ( x  2)10  ...  ( x  100)10

Q.173 lim =
x  x10  1010
(A) 0 (B) 1
(C) 10 (D) 100 [D]
2 2
(a  x ) sin(a  x )  a sin a
Q.174 The value of xlim
0
is
x
2 2
(A) a cos a + 2a sin a (B) a sin a + 2a cos a
(C) a2 cos a + a2 sin a (D) None of these
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 (a  x ) 2 sin(a  x )  a 2 sin a
Sol.[A] Here xlim
0
x
d
= (a2 sin a) = a2 cos a + 2a sin a
da
log x n  [ x ]
Q.175 The value of xlim

, n  N and [.] is G.I.F.
[x]
(A) –1 (B) 0
(C) +1 (D) Does not exist
n n
Sol.[A] lim log x  [ x ] = lim log x  x
x  x 
[x] x
1 n
lim .nx n 1  1 lim 1
 x  x n = x  x =–1
1 1
tan 2 x  n sin x
Q.176 If m, n  I0 and xlim
0 = some integer then value of this limit is
x3
(A) 3 (B) 2
16  n
(C) (D) None of these
12

Sol.[A] lim tan 2 x  n sin x = I

x 0 3
x
8x 3 nx 3
2 x  .....  nx 
 xlim
0 3! 3! = I
3
x
 16  n  3
lim (2  n ) x    x  .....
 x 0  6  =I
x3
16  n
 n = 2 and value = =3
6
n n
1
Q.177 If  t r  2(3n  1) ,  n  1 then nlim
  tr
is equal to
r 1 r 1

(A) 3 (B) 3/2 (C) 3/4 (D) 3/8

n

Sol. [D]  t r = 2(3 n

– 1)
r 1
 t1 + t2 + …..+ tn = 2(3n – 1)
We can do tn = Sn – Sn–1
 tn = 2(3n –1) – 2(3n–1 – 1)
= 2[3n – 3n–1] = 4.3n–1 Now,
n 1
1 1  1  1 3
lim
n   4.3n 1 = . 1   ..... =
4  3
. 
 4 1  
1 =
8
r 1
 3
12  2 2  32  .....  n 2
Q.178 Let  = nlim
 3
&
n

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 (13 – 12 )  ( 23 – 2 2 )  .......  (n 3 – n 2 )
= nlim
 then-
n4
(A)  =  (B)  < 
(C) 4 – 3 = 0 (D) 3 – 4 = 0

12  2 2  32  ......  n 2 1
Sol.[D]  = nlim
 =
3
n 3
3 3 3
1  2  .....  n 1  2 2  .....  n 2
2
 = nlim
 4
–
n n4
1 1
= –0=
4 4
1 1
 3 – 4 = 3 × –4× =1–1=0
3 4
Q.179 The value of the xlim
0
(1 + sin x)2cotx is -

(A) e3 (B) e
2
(C) e (D) None of these

Sol.[C] xlim
0
(1+sin x)2cot x (1 form)
lim cos x
= e x0 (1  sin x –1)  2cot x = lim 2 sin x 
sin x
e x 0
= e xlim
0 2 cos x = e
2

Q.180 lim sin x =

x 
x
(A) 1 (B) 2
(C) 0 (D)does not exist
h sin x sin  [–1to1]
Sol.[C] x 
= = =0
x  

Q.181 lim sin[cos x ] ( [ ] denotes greatest integer function ) equal -

x 0 1  [cos x ]

(A) 1 (B) 0
(C) does not exist (D) None of these
Sol. [B] (i) If x = Rational No. then h n . x = Integer 
h 
 cos ( |n . x) = (–1 or 1)
cos2n ( |n x) = + 1
h (1 + cos2m ( |n x)) = 1 + 1 = 2
n  m

h |n x= (Non Integer)

(ii) If x Irrational No then n 
 – 1 < cos ( |n x) < 1
h
 m 
n 
(cos2m( |n x)) = 0  (less than 1) = 0
h
 m 
n 
(1 + cos2m |n x) =1 + 0 = 1

1 – tan x      
Q.182 Let f(x) =
4x – 
, x  ,x 0, 2  . If f(x) is continuous in 0, 2  , then f  4  is -
4      
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 1 1
(A) – 1 (B) (C) – (D) 1
2 2
 h 1 – tan x 0
Sol. [C] f   = x  / 4
= from
4 4x –  0
– sec 2 x
h – sec 2  / 4
Dh = x  / 4
=
4 4
2
– sec  / 4 2 1
= = – =–
4 4 2
x100  x 50  50
Q.183 lim is
x  x100  x
(A) 100 (B) 10
(C) 1 (D) None of these
Sol.[C] Degree (Num.) = degree (den.)
Ans = 1
1/ n
 en 
Q.184 lim   is
n   
 
(A) 1 (B) e
(C) e2 (D) e
1
 en n
Sol.[B] y  lim  
n    
 

1.  en 
 y  lim log 
n  n   
 
log y = 1  y =e
x4
Q.185 lim  x  6  =
x 
 x 1 
(A) e (B) e4 (C) e5 (D) e6
x4
Sol.[C] lim  x  6  = 1 form
x 
 x 1 
lim ( x  4)  x 6 –1 lim ( x  4)   ( x 6) –( x 1) 
  x 1  = x  x 1 
e x e  

5 x  20
= lim = e5
e x x 1

Q.186 lim {x} sin( x – 2) =

x 2 2
( x – 2)
(A) 0 (B) 
(C) 1 (D) does not exist
Sol.[C] x 2  x = 2 + h
+

sin( 2  h – 2)
limit = hlim {2 + h}
0 ( 2  h – 2) 2
sin h
= hlim
0
h× = 1.
h2
Q. 187 If [x] denotes the greatest integer  x, then

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 1
lim 2 2 2 2
n 3 {[1 x] + [2 x] + [3 x] + ….+ [n x]} equals :
n 

(A) x/2 (B) x/3

(C) x/6 (D) 0
Sol. [B] xI, [x] = x
12 x  2 2 x  .....  n 2 x x (12  2 2  .....  n 2 )
Lim  Lim
n  n3 n  n3
1
= x. (difference of power of Nr and Nr is one)
3
x 2  9 x  20 x 2  9 x  20 x 2  9 x  20 x 2  9 x  20
Q. 188 If P = lim  lim and Q = lim  lim ,
x 5 x  [x] x 4 x  [x] x 4 x  [x ] x 5 x  [x]
[] = G.I.F., then P/Q =
(A) 1 (B) 2
(C) 3 (D) None
x 2  9 x  20 x 2  9x  20
Sol. [D] P = Lim  Lim
x 5 x  [x] x 4 x  [x]
( x  4) ( x  5) ( x  4) ( x  5)
= Lim  Lim
x 5 x 5 x 4 x 3
=1–0=1
x 2  9 x  20 x 2  9 x  20
Q = Lim  Lim
x 4 x  [x] x 5 x  [x]
( x  4) ( x  5) ( x  4) ( x  5)
= Lim  Lim
x 4 
x4 x 5 x4
= – 1 – 0 = – 1, P/Q = – 1

1 
Q.189  2 (1  cos 2 x ) is
 
lim
x 0 x
(A) 1 (B) –1
(C) does not exist (D) None of these [C]

f ( x )e nx  g ( x )
Q.190 If x > 0 and g is a bounded function, then lim is
n  e nx  1
(A) 0 (B) f(x)
(C) g(x) (D) None [B]
x.2 x  x
Q.191 lim equals
x  0 1  cos x

1
(A) log 2 (B) log 2
2
(C) 2 log 2 (D) None [C]

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Q.192 lim x cos    sin    is -
x 
 4x   4x 
 
(A) (B)
2 4
(C) 1 (D) None of these [B]
lim 1  cot 3 x
Q.193  is -
x
4 2  cot x  cot 3 x
11 3
(A) (B)
4 4
1
(C) (D) None of these [B]
2

Q.194 lim 1  x 4  (1  x 2 ) is –
x 
x2
(A) 0 (B) – 1 (C) 2 (D) – 2 [A]

Q.195 lim  3x 
 9x 2  x 
 equals?
x    

1 1 1 1
(A) (B) (C) – (D) – [B]
3 6 6 3

lim n!
Q.196 is equal to -
n  ( n  1) !  n !
(A) 0 (B) 
(C) 1 (D) None of these [A]

Q.197 The value of xlim


x1/x is -

(A) 0 (B) 1
(C)  (D) None of these [B]
 x3 1 
Q.198 If xlim 
  x 2  1
 (ax  b)  = 2, then -
 
(A) a = 1, b = 1 (B) a = 1, b = 2
(C) a = 1, b = – 2 (D) None of these [C]
log( x  a )
Q.199 The value of lim is -
x a log(e x  e a )
(A) 1 (B) – 1
(C) 0 (D) None of these [A]
3
lim x 2 1  x3 1
Q.200 x  4
equals -
5
x4 1  x4 1
(A) 1 (B) 0
(C) – 1 (D) None of these [B]
x2
lim e  cos x
Q.201 x 0
is -
x2

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 3 1
(A) (B)
2 2
2
(C) (D) None of these [A]
3
sin px
Q.202 If xlim
0 tan 3x
= 4, then p is equal to -

(A) 6 (B) 9 (C) 12 (D) 4 [C]

 x 3 
Q.203 The value of xlim  log a
3 
 is -
 x  6  3 
(A) loga 6 (B) loga 3
(C) loga 2 (D) None of these [A]

Q.204 lim ([x – 3] + [3 – x] –x), where [.] denotes the greatest integer function, is equal to -
x 3

(A) 4 (B) – 4
(C) 0 (D) None of these [B]
 4 
Q.205 The value of xlim

3x sin  x  is -
3 
(A) 4 log 3 (B) 3 log 4
(C) 4 (D) None of these [C]

Q.206 If xlim
0
(1 + ax)b/x = e2, (a, b  N), then -

(A) a = 4, b = 2 (B) a = 8, b = 4
(C) a = 16, b = 8 (D) None of these [D]
 sin(1  [ x ])
 , [x]  0
Q.207 If f(x) =  [x] then lim f(x) is -
x 0

 0, [x]  0
(A) – 1 (B) 0
(C) 1 (D) None of these [B]

Q.208 lim log (1 / 4 1 / 8 1 / 16 ......to n terms)

is equal to -
n  0.2 5

(A) 2 (B) 4 (C) 8 (D) 0 [B]

Q.209 If f(x) = [2x], then -

(A) lim f(x) = 0

x 1/ 2 

(B) lim f(x) = 1

x 3 / 4 

(C) xlim
1 / 2
f(x) = 0

(D) x lim
3 / 4
f(x) = 2 [B]

Q.210 The value of

 3 1
 x  2x 
1 
 x  4 x  2  is -
lim    
x  2  x 3  8   x2 x 2  
   

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 (A) 1/2 (B) 2
(C) 1 (D) None of these [A]

Q.211 lim [log n – 1(n).log n(n + 1).log n + 1

n 

(n + 2)… log n k 1 (nk)] is equal to -

(A)  (B) n
(C) k (D) None of these [C]
n
1  2  32  ...  r 2
2 2
Q.212 If tr =
13  2 3  33  ....  r 3
and Sn = 
r 1

(–1)r.tr, then nlim S is given by -

 n

2 2 1 1
(A) (B) – (C) (D) – [B]
3 3 3 3
x x  x 
Q.213 The value of nlim

cos   cos   …cos  n  is-
2
  4
  2 
sin x
(A) 1 (B)
x
x
(C) (D) None of these [B]
sin x

Q.214 lim (1+ x) (1 + x2) (1 + x4) ….. (1 + x2n), | x | < 1, is equal to -

n 

1 1
(A) (B)
x 1 1 x
(C) 1 – x (D) x – 1 [B]
2 2 2 2
Q.215 lim tan[e ]x  tan[ e ]x is,
x 0
sin 2 x
(where [.] is G.I.F.) -
(A) 0 (B) 8
(C) 15 (D) None of these [C]
e x  e tan x
Q.216 The value of xlim
0
equals -
x  tan x
(A) 0 (B) 1 (C)  (D) ½
 x  tan x
 1  tan x 
e .
Sol.[B] xlim  x  tan x  e =1
0   
    
1 1

f (x)  1
Q.217 If f(1) = 1, f (1) = 2 then lim
x 1 =
x 1
(A) 2 (B) 1 (C) 3 (D) 4
f (x)  1 0
Sol.[A] lim
x 1
; form
x 1 0

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 1
.f ' ( x )
2 f (x) f ' (1)
 lim
x 1 = . 1 =2
1 f (1)
2 x
1
2 2 1
Q.218 lim (a  h ) . sin (a  h )  a sin a equals -
h 0
h
(A) a2 cos a + 2a sina
(B) – a2 cos a – 2a sin a
a2
(C) + 2a sin–1 a
1 a2
(D) none of these
d
Sol.[C] The limit equals (a2 sin–1 a)
da
a2
 + 2a sin–1 a
1 a2

lim 1  cos x 2
Q.219 x 0
equals -
(1  cos x )

1
(A) 2 (B)
2
(C) 1 (D) none of these

x4
11 .....
Sol.[A] xlim 2! = 2
0
x2
11 .....
2!
x.e1/ x
Q.220 The value of xlim
0 
is -
1  e1/ x
(A) 0 (B) 1
(C)  (D) none of these

lim  x.e
1/ x 

Sol.[A] As we are RHL so x 0  1  e1/ x 
 
 1 
x.e1/ x  2 .  e1/ x
lim x 
x 0
 1 
= e1/ x  2  =0
x 
Q.221 lim  4  1  equals to -
x 2
 x2  4 2  x 

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 1 1 3 3
(A) – (B) (C) – (D)
4 4 4 4

lim  4  1  lim  4  x  2  lim  ( x  2) 1

Sol.[A] x 2
 x 2  4 x  2  = x 2  x 2  4  = x 2 ( x  2) ( x  2) = 4
sin 1 x  tan 1 x
Q.222 xlim
0
equals -
x3
(A) 1 (B) –1 (C) 1/2 (D) –3/2
Sol.[C] Hint : First once L' Hospital then rationalization
(1  x )1/ x  e
Q.223 The limit, xlim
0
equals -
x
(A) e (B) e/2 (C) – e (D) – e/2
Sol.[D] y = (1 + x) 1/x
 log y = 1/x log (1 + x)
 x  x
 1 ...   ...
 y = e 2 
 e.e 2
  x ..... 
lim e e 2  1 e
 
x 0   
so 2
x
2x
 x 
Q.224 If f (x) =   then -
2x 

(A) xlim

f (x) = e–6 (B) xlim

f (x) = 2

(C) xlim

f (x) = e–3 (D) xlim

f (x) = e–4
2x 2x
lim  x   2 
lim  2  x 
Sol.[D] x   2  x 
; 1 form  e = e–4 x   

Q.225 lim  1  1  1  ...  1 

 equals -
n   2.3 3.4 4.5 n.( n  1) 

(A) 1 (B) 0 (C) 1/2 (D) 2

lim  1  1  1  1  ...  1  1  1
Sol.[C] n  2 3 3 4 n n  1  
  2
sin x
sin x  (sin x )
Q.226 lim
 1  sin x  ln sin x equals -
x
2
(A) 1 (B) 2
(C) 3 (D) 4
t  tt 0
Sol.[B] Put t = sin x  lim ,
t 1 1  t  log t 0
1  t t (1  log t )
lim 0
t 1 1 ,
0 1  0
t
 1
0  t t  0    (1  log t ).(t t (1  log t ))
 t 1  1
lim = =2
t 1 1 1
0 2
t

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 1
Q. 227 The value of
(1  cos 2 x ) is -
2
lim
x 0 x
(A) 1 (B) –1
(C) 0 (D) None of these
1
 2 sin 2 x | sin x |
Sol.[D] 2  lim = does not exist
lim x0 x
x 0 x
x  sin x n
n
Q. 228 If lim is nonzero finite, then n may be equal to
x 0 x  sin n x
(A) 1 (B) 2
(C) 3 (D) None of these
Sol.[A] Clearly by option at x = 1
x´ sin x´
lim = 1 (non zero and finite)
x 0 x  sin´x
tan(ax 2  bx  c)
Q.229 If  is a repeated root of ax2 + bx + c = 0, then lim is -
x  ( x  ) 2
(A) a (B) b
(C) c (D) 0
tan ( x   ) 2
Sol.[A] lim =a
x  ( x  ) 2
log e [ x ]
Q.230 lim , where [x] denotes the greatest integer less than or equal to x, is -
x  x
(A) 1 (B) –1
(C) 0 (D) does not exist
x log x 1/ x
Sol.[C] [ x ]  x xlim
 x
= lim
x  1
=0

y3
Q.231 lim as (x, y)  (1, 0) along the line y = x – 1 is given by
x 1 x3  y2 1
(A) 1 (B) 
(C) 0 (D) none of these
y3
Sol.[C] lim
x 1 x 3  y 2  1 , y = x – 1
y 1

( x  1) 3 0
lim 3 2 ,
x 1 x  ( x  1)  1 0
3( x  1) 2
lim =0
x 1 3x 2  2( x  1)

Q. 232 If a = min {x2 + 4x + 5, x  R} and

n
1  cos 2
b = lim
 0  2
then the value of a
r 0
r
.b n  r is-

n 1
2 1
(A) (B) 2n + 1 – 1
n
4 .2
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 2 n 1  1
(C) (D) none of these
3 .2 n
4.1.5  4 2
Sol.[D] a = min{x2 + 4x + 5} = =1
4 1
1  cos 2
b = lim =2
 0 2
n n

r 0
a r .b n  r = 1.2
r 0
n r

= 2n(1 + 2–1 + 2–2 + .... + 2–n)

= 2n 1

 1  1 / 2  n 1   = 2 n+1
–1
1  (1 / 2) 
 
3 f (x)
Q.233 If f(9) = 9 and f ´(9) = 1, then lim is equal to -
x 9 3 x
(A) 0 (B) 1
(C) –1 (D) None of these
1f ´(x )
0
3 f (x) 0 2 f (x)
Sol.[B] lim ,  lim
x 9 3 x 0 x 9 1
0
2 x
f ´(9) 1 3  2
=+ × 2 9 = =1
2 f (9) 23
e x  e tan x
Q.234 The value of xlim
0
equals -
x  tan x
(A) 0 (B) 1 (C)  (D) ½
 e x  tan x  1 
 . tan x
Sol.[B] xlim  x  tan x  e =1
0 
      
1 1

f (x)  1
Q.235 If f(1) = 1, f (1) = 2 then lim
x 1 =
x 1
(A) 2 (B) 1 (C) 3 (D) 4
f (x)  1 0
Sol.[A] lim
x 1
; form
x 1 0
1
.f ' ( x )
2 f (x) f ' (1)
 lim
x 1 = . 1 =2
1 f (1)
2 x
1 2 2 1
Q.236 lim (a  h ) . sin (a  h )  a sin a equals -
h 0
h
(A) a2 cos a + 2a sina
(B) – a2 cos a – 2a sin a

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 a2
(C) + 2a sin–1 a
1 a2
(D) none of these
d
Sol.[C] The limit equals (a2 sin–1 a)
da
a2
 + 2a sin–1 a
1 a2

lim 1  cos x 2
Q.237 x 0
equals -
(1  cos x )

1
(A) 2 (B)
2
(C) 1 (D) none of these

x4
11 .....
Sol.[A] xlim 2! = 2
0
x2
11 .....
2!
x.e1/ x
Q.238 The value of xlim
0 
is -
1  e1/ x
(A) 0 (B) 1
(C)  (D) none of these

lim  x.e
1/ x 

Sol.[A] As we are RHL so x 0  1  e1/ x 
 
 1 
x.e1/ x  2 .  e1/ x
lim x 
x 0
 1 
= e1/ x  2  =0
x 
Q.239 lim  4  1  equals to -
x 2
 x2  4 2  x 
1 1 3 3
(A) – (B) (C) – (D)
4 4 4 4

lim  4  1  lim  4  x  2 
x 2
Sol.[A]  x 2  4 x  2  = x 2  x 2  4 
lim  ( x  2) 1
= x  2 ( x  2) ( x  2) = 4
sin 1 x  tan 1 x
Q.240 xlim
0
equals -
x3
(A) 1 (B) –1 (C) 1/2 (D) –3/2
Sol.[C] Hint : First once L' Hospital then rationalization
(1  x )1/ x  e
Q.241 The limit, xlim
0
equals -
x
(A) e (B) e/2 (C) – e (D) – e/2
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Sol.[D] y = (1 + x)1/x  log y = 1/x log (1 + x)
 x  x
 1 ...   ...
y= e 2   e.e 2
  x ..... 
lim e e 2  1 e
 
x 0   
so 2
x
2x
 x 
Q.242 If f (x) =   then -
2x 

(A) xlim

f (x) = e–6 (B) xlim

f (x) = 2

(C) xlim

f (x) = e–3 (D) xlim

f (x) = e–4
2x 2x
lim  x  lim 
 2 

Sol.[D] x   2  x 
; 1 form  e 
= e–4 x   2  x 

Q.243 lim  1  1  1  ...  1 

 equals -
n   2.3 3.4 4.5 n.( n  1) 

(A) 1 (B) 0 (C) 1/2 (D) 2

Sol.[C] lim  1  1  1  1  ...  1  1   1

n 2 3 3 4 n n  1 
 2
 x2 
 
lim
Q.244 The value of x 0 


0
sec 2 t dt 

is :
 x sin x 

(A) 3 (B) 2
(C) 1 (D) 0
x2

 sec
2 x2
t dt
 sec
2
t dt
Sol.[C] xlim
0
0  xlim
0 0
 sin x 
x x x2
 x 
sin x
 xlim
0
=1
x
Apply L-Hospital rule
2 2
lim sec x .2 x
x 0
2x
2
sec 0 = 1

lim  1  2  ......  n  =
2 2 2
Q.245 n   3 3 3 
n n n 
1 1
(A) (B)
6 3
2
(C) (D) None of these
3
2 2 2
Sol.[B] lim 1  2  .....  n
n 
n3

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 lim n ( n  1)(2n  1)
n 
6n 3
n 3 [1  1 / n ][2  1 / n ] 2 1
 nlim
 = =
6n 3 6 3
lim sin x  x
Q.246 x0 =
x3
1 1 1 1
(A) (B) – (C) (D) –
3 3 6 6
lim sin x  x  0 
Sol.[D] x  0 3  form 
x 0 
L. Hospital rule

lim cos x  1 lim  2 sin 2 ( x / 2)

x0 = x 0
3x 2 4  ( x / 2) 2  3

2 1
= = 
3 4 6
Q.247 If 1 + sinx + sin2 x + ……to  =4+2 3

0 < x <  then :

 
(A) x = (B) x =
6 3
   2
(C) x = or (D) x = or
3 6 3 3
lim x
Q.248 x    x  3  equals :
 x2
(A) e (B) e–1 (C) e–5 (D) e5
Sol.[C] Given limit is ( 1 form)

lim x  x  3 1
limit =
e x   x2 

lim x  5 
= = e–5
e x   x 2 

lim 1 – cot 3 x
Q.249 x  / 4
is -
2 – cot x – cot 3 x
(A) 11/4 (B) 3/4
(C) 1/2 (D) None of these
h (1 – cot x ) (1  cot x  cot 2 x ) 1  1  1 3

Sol.[B] x  4 (1 – cot x ) (cot 2 x  cot x  2) = 2  1  1 = 4 .
1/ 2 1/ 3
Q.250 lim (cos x ) – (cos x ) is -
x 0
sin 2 x
(A) 1/6 (B) – 1/12 (C) 2/3 (D) 1/3

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 1 1
(cos x ) 2 – (cos x ) 3  
h 0
x  0
Sol. [B] 0
sin 2 x
1 2
– 1 –
h 1 (cos x ) sin x  (cos x ) 3 sin x
2
x 0 3
 – 2
2 sin x. cos x
1 2
2  1 (cos x ) 3
– –
h 1 (cos x )
x 0 3
  – 2
2 cos x
1 1 1
–
= – 2  2 3 
1
= – 12
x
 x 2  5x  3 
Q.251 If f(x) =  2 
 then xlim

f(x) is -
 x  x  2 
(A) e–4 (B) e3 (C) e2 (D) e4
x
 x 2  5x  3 
h
 
Sol. [D]  x   x 2  x  2   1 form
Using f9 = e9(f – 1)
h  x 2  5x  3  h (4 x  1) x
e x   – 1 e x 
 x.  x 2  x  2  = x2  x  2
41
 1 = e4
e

lim ax –1
Q.252 x 0
is -
ax – a

(A) 2 a log a (B) a log a

(C) log a (D) none of these

a x log a
h ax –1 0 h h
x 0
  x 0 1 1 ax
Sol. [A] ax – a 0  = x 0 2axlog a ×
2 ax
=2 a log a

lim sin( x 1 / 3 )n (1  3x )

Q.253 x 0 1/ 3 =
(tan 1 x ) 2 (e 5 x  1)
(A) 3/5 (B) 1/5
(C) 2/5 (D) None of these
 sin x1 / 3  1/ 3   log (1  3x ) 
 1/ 3
x  
  (3x )
 x    (3x ) 
Sol.[A] xlim
0 2
 tan 1 x   5x  1 
1/ 3

  . x . e . 5x1 / 3
 x   5x1 / 3 
   
= 3/5

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 a b
Q.254 If lim x sin x , a, b, c  R – {0}, exists and
x 0 c
sin( x )
non-zero, then-
(A) a – b + c = 0 (B) a – b – c = 0
(C) a + b – c = 0 (D) None of these
b
a  sin x
 b
x   .x
 x 
Sol. [C] xlim
0  sin x c 
  c
 xc  . x
 
lim xa + b – c
x 0
For non zero value a + b – c = 0

Q.255 lim sin || x | 2 | 3 is-

x 1

(A) sin 2 (B) sin 1

(C) 0 (D) Does not exist

Sol. [A] R.H.L. hlim

0
sin ||1 + h –2| –3|

lim sin | 1 – h –3|  lim sin (2 + h) = sin 2

h 0 h 0
L.H.L.
lim sin |1 (1 – h) –2| –3|
h 0

lim sin |1 + h –3|  lim sin (2 – h) = sin 2

h 0 h 0

Q.256 lim |x| sin x =

x 0

(A) 0 (B) does not exist

(C) 1 (D) None of these

Sol. [C] Let y = xlim

0
|x|sinx 00 form

log y = xlim
0
sin x log |x| 0 × 
log | x | 
log y = xlim
 0 cosec x 

Apply L' Hospital rule

1/ x
log y = xlim
0
cosec x cot x
 sin x 
log y = xlim
0
–   tan x
 x 
log y = 0
y=1

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Q.257 If x > 0 and g is a bounded function, then

nx
lim f ( x )e  g ( x ) is-
n 
e nx  1
(A) 0 (B) f(x)
(C) g(x) (D) None of these
Sol. [B] g is a bounded function
 value of g is finite.

 g( x )  g( x )
e nx f ( x )  nx  f (x) 
lim  = f(x)
n  
 e  =
1
e nx [1  1 / e nx } 1


Q.258 lim  1 
4

9
 ..... 
n2 
=
n  3
n3  1 n3  1 n 3  1 
 n  1
(A) 1 (B) 2/3
(C) 1/3 (D) 0
12  2 2  32  ......  n 2
Sol. [C] nlim

1 n3
lim n ( n  1) (2n  1)
n  6 (1  n 3 )
3
lim n (1  1 / n ) (2  1 / n ) = 1. 2 = 1
n  3 3
6n (1 / n  1) 6 3

Q.259 Let f(x) = x(–1)[1/x], x  0, where [x] denotes the greatest integer less than or equal to x then, xlim
 0 f(x) =

(A) doesn't exist (B) 2

(C) 0 (D) –1
Sol. [C]  [1/x] = Integer
 (–1)[1/x] 1 or –1
lim
x 0 x (–1)[1/x]
lim (h ) (1 or  1)  0 
h 0 
lim (h ) (1 or  1)  0
h 0 
lim  (cos 1 x )
Q.260 x  1
is given by-
( x  1)

(A) 1/  (B) 1 / ( 2)

(C) 1 (D) 0
–1
Sol. [B] Let cos x = y
 x  –1, y  cos–1 (–1) y 
  y  y 
lim 
y 1  cos y   y 

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  y
 
lim y put y =  + h
y
2 . cos  y
2
h
 h
  sin  1
lim
h  2. 
h
2
  h . h =
2
2 
 
2
a 2 g(x). g(1/x) = g(x) + g(1/x)  g(x) = xn + 1
Q.261 lim n sin ( n!) , (0 < a < 1) =
n 
n 1 g(2) = 2n + 1 5 = 2n + 1
 n = 2  g(n) = n2 + 1
(A) 1 (B) +
(C) 0 (D) None of these lim (x2 +1) = 32 + 1 = 10
x 3

n a 1  sin 2 ( n!) Q.264 If f(x) = sin [x]/[x], [x]  0

( ) a 1 .[0, 1]
Sol. [C] nlim
  1 = = 0, [x] = 0
1   1 where [x] denotes the greatest integer less than
 n
= 0 × [0, 1] or equal to x; then xlim
0
f(x) equals-
( 0 < a < 1 a–1 –ve = 0) = 0
(A) 1 (B) 0
Q.262 lim log (1 / 4  1 / 8  1 / 16 ...n terms )
is (C) –1 (D) None of these
n  (0.2)
5

equal to-  sin[ x ]

, x  [0, 1)
Sol. [D]  [ x ]
(A) 2 (B) 4 
 0, x  [0, 1)
(C) 8 (D) 0
  1   RHL lim 0 = 0
   1 n   x 0 
1
log 5     
2
Sol. [B] nlim
 4 1  = lim sin [ x ]
  1   LHL
  2   x 0  [x]
(0.2)
1 1 
lim sin [  h ] = sin [ 1] = sin 1
1   h 0 [h ] [ 1]
5 2
log
( 0 .2 )    
Limit does not exist.
1 1
log   1 n
= 1
5 2
=
 a k and nlim
  log 5  
5 5 2 Q.265 If Sn = 
a n = a, then
k 1
log  2  1
= 5 5
5 1/ 2
log5 2  5 2 log5 2 =
(S n 1  S n )
log 5 ( 2) 2 =4
5 lim n is equal to-
Q.263 If g(x) is a polynomial satisfying
n 
 k
g(x) g(y) = g(x) + g(y) + g(xy) –2 for all real x k 1

(A) 0 (B) a (C) 2 a (D) 2a

and y and g(2) = 5, then xlim
3
g(x) is-
Sol. [A] Sn = a1 + a2 + …..+ an
(A) –8 (B) 10 Sn+1 = a1 + a2 +……+ an + an+1  Sn+1 –Sn = an+1
(C) 8 (D) None of these
Sol. [B] Put x = 2, y = 1
a n 1
lim n (n  1)  nlim a =a
g (2) g(1) = g(2) + g(1) + g(2) –2 n   n
4g (1) = 8 (g(2) = 5) 2
g(1) = 2
 nlim
 an+1 = a
Put y = 1/x

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 a (A) 1 (B) 2/3 (C) 1/3 (D) 0 [C]
= =0

x 1
Q.266 If root of equation ax2 + bx + c = 0 is x =  and
Q.272 The value of xlim  3x  4 
  
3
is equal to
lim  3x  2 
root's are repeated then value of x 
(A) e–1/3 (B) e–2/3 (C) e–1 (D) e–2 [B]
2
sin(ax  bx  c) Q.273 The value of
2 is-
( x  ) lim cos  x  cos  x  cos  x  …cos  x 
n 
(A) a (B) –a 2 4 8  2n 
(C) 0 (D) None of these is
Sol. [A]  root's are repeated.
sin x
 ax2 + bx + c = a(x – )2 (A) 1 (B)
x

sin a ( x   ) 2  x
 xlim
  2  a=a (C) (D) None of these [B]
 a ( x  )  sin x
(1  cos 2 x ) sin 5x Q.274 Let f : R  R be a differentiable function
Q. 267 The value of xlim
0 2
is-
x sin 3x  1 
having f (2) = 6, f (2) =   . Then
(A) 10/3 (B) 3/10 (C) 6/5 (D) 5/6 [A]  48 
lim sin (cx  bx  a )
2
Q. 268 The value of x  1 f (x)
4t 3
 x  1
(where  and  are roots of ax2 + bx + c = 0) is-
lim
x 2 
6
x2
dt equals

  (A) 12 (B) 18 (C) 24 (D) 36 [B]

(A) c   (B)  
      sin x cos x tan x
1 c  1 1 Q.275 If f (x) = x3 x2 x , then
(C) (D)    [D] 2x 1 1
   
3
lim x2  2 3
x 1 lim f ( x ) is
Q. 269 x 1
is equal to- x 0
2
( x  1) x2
1 1 (A) 3 (B) –1 (C) 0 (D) 1 [D]
(A) (B) n
9 6
1 Q.276 If Sn = a k and lim an = a, then
n 
(C) (D) None of these [A] k 1
3
Q.270 Let the function f be defined by the equation S n 1  S n
if 0  x  1 lim n
is equal to
 3x
f (x) = 
5  3x if 1  x  2
, then
n 
k k 1

(A) lim f ( x )  f (1) (A) 0 (B) a (C) 2 a (D) 2a [A]

x 1
Q.277 Let f (a) = g (a) = k and their nth derivatives
(B) lim f (x)  3 f n (a), g n (a) exist and are not equal for some n.
x 1
If lim
(C) lim f (x)  2 x a
x 1
f (a )g ( x )  f (a )  g (a )f ( x )  g (a )
(D) lim f ( x ) does not exist [D] =
x1 g( x )  f ( x )
4, then the value of k is
Q.271 lim  1 
4

9
 ... 
n2 
 (A) 4 (B) 2 (C) 1 (D) 0 [A]
n  3
n3 1 n3 1 n 3  1 
 n  1
=

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 2x a xy
 a b 
If xlim
 e
2
1   2  = e2, then the values sin 2 t
  e sin
Q.278 t
 dt
 x x  Sol.[A]
y a
of a and b are lim
x 0 x
(A) a  R, b  R (B) a = 1, b  R x y
2

(C) a  R, b = 2 (D) a = 1, b = 2 [B]  e sin t dt

= e y

Q.279 The value of lim lim

x 0 x 0 x
sin 2 ( x  y )
 e x log( 2 x 1)  ( 2 x  1) x sin x 
1/ x
= lim e .1  0 2
  is equal to  e sin y

 e x log x  x 0 1
 
1
(A) e (B) log 2
e
sin x , x  n, n  Z
(C) e log 2 (D) None [B] Q.285 If f(x) = 
lim sin(cx 2  bx  a )  2, otherwise
Q.280 The value of x
1 x 2  1 , x  0, 2
 x  1 
g(x) = 4 , x  0, then
(where  and  are the roots of ax2 + bx + c = 0)  5 , x2
is 
  lim g (f ( x )) is
(A) c   (B)   x0
     
(A) 1 (B) 5 (C) 6 (D) 7
1 c  1 1
(C) (D)    [D] Sol.[A] xlim g
 lim f ( x ) 

      0  x 0 
lim g (sin x )
 1  1   1  x0
1  2 1  2 ...1  2 
lim sin 2 x  1  1
Q.281 lim  2  3   n 
is equal x 0
n 0
 1  1   1   x, x  0
1  1  ...1   
 2  3   n  Q.286 If f(x) =  1, x  0 , then xlim f(x) is
0
to x 2 , x  0

1 1 (A) 0 (B) 1
(A) – (B) (C) 2 (D) – 2 [B]
2 2 (C) 2 (D) Does not exist
Sol.[A] f(0 – h) = hlim 0h  0
0
lim  cos 1 x 2
Q.282 x  1
is given by f(0 + h) = lim (0  h )  0
x 1 h 0
1/ x
1 1   
(A) (B) Q.287 lim tan   x  is
 2 x 0   4 
(C) 1 (D) 0 [B] (A) 1 (B) –1 (C) e2 (D) e
xb 1   
 tan   x  1
Q.283 lim  x  a  is equal to Sol.[C] lim e x  4  
x 
xb x 0
(A) 1 (B) e b–a
(C) e a–b
(D) e b
[C]  
sec 2   x   0
4  2
/4
a a  = esec  e2
1  lim e 1
 
2 2
Q.284 lim  e sin t dt  e sin t dt  is equal to
x 0 x   x 0
y xy  Q.288 If x is real number in [0, 1], then the value of
(where a is a constant) lim lim [1 + cos2m (n! x)] is given by
(A) esin2y (B) sin 2y esin2y m  n 
(C) 0 (D) None of these (A) 2 or 1 according as x is rational or irrational
(B) 1 or 2 according as x is rational or irrational
(C) 1 for all x
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 (D) 2 or 1 for all x. x
(C) (D) None of
Sol.[A] mlim lim 1 + (cos2 n! x)m
 n 
sin x
these
Case I
when x is irrational = 1 + (cos2 ) sin x
x x x
= 1 + (0 to 1) = 1 + 0 = 1 Sol.[B] cos cos 2 ....cos n = n  x 
put x = 0 when x is rational 1 + cos 0 = 2 2 2 2 2 sin  n 
2 
sin(e x 1  1) by trigonometry
Q.289 lim is equal to
x 1 log x sin x
(A) 1 (B) 0
 x  x
(C) e (D) None of these
lim 2 n . sin  n  . n sin x
sin(e x 1  1)
= n  2  2 =
Sol.[A] lim x
x 1 log x  x 
 n
cos(e x 1  1).e x 1 2 
= lim
x 1 1 =1 sin x n
Q.293 For m, n  I+, Lt is equal to -
x x 0 (sin x ) m
 1 (A) 1, if n = m (B) – 1, if n > m
x sin  , x  0
Q.290 If f(x) =  x n
(C) (D) None of these [A]

 0 , x0 m
Q.294 Let f ''(x) be continuous at x = 0 and f '' (0) = 4.
lim f ( x )
Then x 0 Then value of
(A) is equal to 1 (B) is equal to – 1
2f ( x )  3f (2 x )  f ( 4 x )
(C) is equal to 0 (D) does not exist. lim is -
1
x 0 x2
Sol.[C] f(0 + h) = hlim
0
h sin = 0 sin  = 0 (A) 12 (B) 10
h
(C) 6 (D) 4 [A]
 1 
f(0 – h) = hlim
0
– h sin   = 0 sin (–) = 0
h Q.295 The value of lim 1 2  x  3 is -
Q.291 Suppose f : R  R is a differentiable function x 2 x2
and f(1) = 4. Then the value of 1 1
(A) (B)
2t
f (x) 8 3 4 3
lim
x 1  4( x  1)
dt is
(C) 0 (D) None of these [A]
(A) 8 f '(1) (B) 4 f '(1) x  sin x
(C) 2f '(1) (D) f '(1) Q.296 lim is equal to -
f (x)
x 0 x  sin 2 x

Sol.[A] lim
x 1
 2t dt
4
(A) 1
(C) 
(B) 0
(D) None of these [B]
x 1 20

= lim
2 f ( x ) f ( x )  0
= 2f(1) f (1) = 8f Q.297 x
k 1
k
 20
=
x 1 1 0 lim
x 1 x 1
(1)
Q.292 The value of (A) 20 (B) 210
(C) does not exist (D) None of these [B]
lim cos  x  cos  x  cos x
 … cos Q.298 If x1, x2 are real and distinct roots of
n 
2 4 8
ax2 + bx + c = 0, then
 x 
 n  is lim (1  sin (ax 2  bx  c))1 / x  x1 is equal
2  x  x1
sin x to -
(A) 1 (B)
x
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 (A) e x1  x 2 (B)

e x 2  x1
(C) e a ( x 1  x 2 ) (D) e a ( x 2  x1 ) [C]

log e x n  [ x ]
Q.299 lim , n  N, ([x] denotes
x  [x]
greatest integer less than or equal to x) is equal
to -
(A) – 1 (B) 0
(C) 1 (D) does not exist [A]
a tan x  a sin x
Q.300 lim (a > 0) is equal to -
x  0 tan x  sin x

(A) – lna (B) lna

(C) 1 (D) – 1 [B]

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