Limit

If the values of f  x  can be made as close as we like to L by taking

values of x sufficiently close to a (but not necessarily equal to a) then

we write
lim f  x   L … (2.1)
x a

which is read “the limit of f(x) as x approaches a is L” or “f(x)

approaches L as x approaches a”.
The expression in Eq. (2.1) can also be written as f  x   L as x  a .

x3  2 x 2 x  x  2  x 2
2

Let us consider f  x     ; x2

3x  6 3 x  2  3

x2 4
Now lim f  x   lim 
x2 x2 3 3
f x
3.0

2.5

2.0

1.5

1.0

0.5

x
-2 -1 1 2 3

x2
Thus the graph of f is the parabola y  with the point (2,4/3)
3
deleted as shown in the figure. It is geometrically evident that as x
gets closer to 2, f(x) gets closer to 4/3.
In general, if a function f is defined throughout an open interval
containing a real number a, except possibly at a itself we may ask the
following questions:
 1. As x gets closer to a (but x  a ) does the function value f(x) get
closer to some real number L?
2. Can we make the function value f(x) as close to L as desired by
choosing x sufficiently close to a (but x  a )?
If the answers to these questions are yes, we use the notation
lim f  x   L
x a

We always assume that x  a ; that is, the function value f(a) is

completely irrelevant. As we shall see, f(a) may be different from L,
may equal L, or may not exist, depending on the nature of the
function.

If the values of f  x  can be made as close as we like to L by taking

values of x sufficiently close to a (but greater than a) then we write

lim f  x   L … (2.2)
x a 

and the values of f  x  can be made as close as we like to L by taking

values of x sufficiently close to a (but less than a) then we write

lim f  x   L … (2.3)
x a 

Expression (2.2) is read “the limit of f(x) as x approaches a from the

right is L”. Similarly expression (2.3) is read “the limit of f(x) as x
approaches a from the left is L” or “f(x) approaches L as x approaches
a from the left”.
The limit in (2.1) is called a two-sided limit because it requires the
values of f(x) to get closer and closer to L as values of x are taken
from either side of x  a . In general, there is no guarantee that a
function f will have a two-sided limit at a given point a; that is the
values of f(x) may not get closer and closer to any single real number
L as x  a . In this case, we say that lim f  x  does not exist. The
xa

limit of a function f(x) to exist at a point a, the values of f(x) must

approach some real number L as x approaches a, and this number
must be the same regardless of whether x approaches a from the left
or right.
Theorem: Let a and b real numbers, and suppose that
lim f ( x)  L1 and lim g ( x)  L2 , Then
xa xa

(a) lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)  L1  L2

x a x a x a

the limit of a sum is the sum of the limits.

(b) lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)  L1  L2
x a x a x a

the limit of a difference is the difference of the limits.

(c) lim  f ( x) g ( x)  lim f ( x) lim g ( x)  L1 L2
x a x a x a

the limit of a product is the product of the limits.

 f ( x)  lim f ( x) L
(d) lim    x a
 1 , L2  0
xa g ( x )
  limx a
g ( x) L2

the limit of a quotient is the quotient of the limits, provided the

limit
of the denominator is not zero.
(e) lim n f  x   n lim f  x   n L1 , provided L1  0 if n is even
xa xa

the limit of an nth root is the nth root of the limit.

If f  x   k is a constant function in (c) we have

lim  k g ( x)   lim k lim g ( x)  k lim g ( x)  k L2

x a x a x a x a

a constant factor can be moved through a limit symbol.

3
Again, lim  f ( x )    lim f ( x )   L13
3

xa  xa 
this is the extension/application of (c).
Note: The above theorem is also true for one-sided limits as x  a  or x  a 
 Infinite limit
Consider the behavior of f ( x)  1 for values of x near 0. If x values
x
are taken closer and closer to 0 from the right, the values of
f ( x)  1 are positive and increase without bound; and as x values
x
are taken closer and closer to 0 from the left, the values of f ( x)  1
x
are negative and decrease without bound. We describing these
limiting behaviors by writing
lim f ( x)  lim 1    and lim f ( x)  lim 1  
x 0 x 0 x x 0 x 0 x
The expressions lim f ( x)    and lim f ( x)    denote that f(x)
x a x a

increases without bound as x approaches a from left and the right,

respectively. If both are true then we write lim f ( x)    .
xa

Example:
1
f ( x) 
 x  a
2

1
 lim f ( x)  lim  
 x  a
2
x a xa

1
lim f ( x)  lim  
 x  a
2
xa xa

Finally, lim f ( x)   
xa
 f x

x
-2 -1 1 2 3 4

Similarly, the expressions lim f ( x)   and lim f ( x)   denote

x a x a

that f(x) decreases without bound as x approaches a from left and the
right, respectively. If both are true then we write lim f ( x)    .
xa

Example:
1
f ( x)  
 x  a
2

1
 lim f ( x)  lim  
 x  a
2
x a xa

1
lim f ( x)  lim  
 x  a
2
xa xa

Finally, lim f ( x)   
xa
f x
x
-2 -1 1 2 3 4

-2

-4

-6

-8
 Limits at infinity
If the values of a variable x increase without bound, then we write
x   , and if the values of x decrease without bound, then we write
1 1
x    . For example lim  0 and lim  0
x  x x x

If the values of f(x) eventually get as close as we like to a number L

as x increases without bound, then we write
lim f  x   L or f  x   L as x   
x 

Similarly, if the values of f(x) eventually get as close as we like to a

number L as x decreases without bound, then we write
lim f  x   L or f  x   L as x  
x

f x

x
-4 -2 2 4

-1

-2
 Infinite limits at infinity
If the values f(x) increase without bound as x   or x    , then
we write lim f  x    or lim f  x   as appropriate
x  x

and if the values f(x) decrease without bound as x   or x    ,

then we write lim f  x     or lim f  x     as appropriate
x  x

Example:
lim x 3  , lim x 3   
x x 

lim x 4  , lim x 4  
x x  

The behavior of a function f(x) as x increases without bound or

decreases without bound is called the end behavior of the function.
1
For example, lim 0
x x
 Finding the Limiting Values

x3  8  x  2  x2  2x  4
lim  lim  lim  x 2  2 x  4   12
x 2 x  2 x 2  x  2 x 2

t  2, t  0

Let g  t    t 2 , 0  t  2 . Find (i) lim g  t  , (ii) lim g  t  , (iii) lim g  t 
t 0 t 1 t 2
 2t , t  2


lim 2
x2  6x  5
 lim 2
x2  5x  x  5
 lim
 x  1 x  5  lim  x  5    4
x 1 x  3 x  4 x 1 x  4 x  x  4 x 1  x  4  x  1 x 1  x  4  5

lim
x4  1
 lim
 x 2  1 x 2  1
 lim
 x 2  1  x  1 x  1
x 1 x  1 x 1 x 1 x 1  x  1
 lim  x 2  1  x  1  4
x 1

1 1 1
lim  lim  lim 
x 2 2  x x2   2  x  x2 x  2

x 2
 
lim 1  x  2  lim 1  lim x  2  1  0  1
x 2 x 2
 sin x cos x
lim  lim  lim cos x  1
x 0 x x 0 1 x 0

lim
1  cos x
 lim
1  cos x 1  cos x 
 lim
1  cos 2 x 
 lim
sin 2 x
x 0 x x 0 x 1  cos x  x 0 x 1  cos x  x 0 x 1  cos x 

 sin 2 x x   sin 2 x   x   0 
 lim  2    lim  lim   1 0
x 0
 x 1  cos x    x 0 x 2
  x 0 1  cos x
  1  1 

tan x  sin x 1   sin x   1 

lim  lim    lim   lim    1 1  1
x 0 x x  0
 x cos x  x  0
 x  x  0
 cos x 

sin 2 sin 2 sin 2

lim  2lim  2 lim 2
 0   0 2 2  0 2

sin 3  3 
 sin 3 3   sin 3   5 
lim  lim    lim  lim
  0 sin 5
 0 sin 5  0
 3 sin 5   0  3   
 5 
 
   
1 1  1 3
 3 lim   3  3   5
5  0  sin 5  5  sin 5  5  1
 5   lim  
  0  5 

cos x  2 x  1  sin x  2 2
lim  lim 
x 0 3x x 0 3 3

4 tan x 4sec 2 x sec x 1

lim  lim  4 lim  4 lim 4
 1  sec x  sec x tan x  tan x  sin x
x x x x
2 2 2 2

1
ln x x  2lim 1  2  0   0
lim  lim
x  x x 1 x  x
2 x
 e3 x 3e3 x 9e 3 x
lim 2  lim  lim 
x  x x  2 x x  2

e x  e x e x 1  e 2 x  1  e 2 x 1  0
lim  lim  lim  1
x  e x  e  x

x  e x 1  e 2 x
 x 1  e2 x 1  0

 2n1   2n1    2  n1 

n 1
3  n1  1 3  n 1  1 lim     1
n   3 
2n1  3n1  3   3     
lim n n  lim  lim 3
n  2  3 n  2 n
 n  2 n
  2  n

3n  n  1  n  1 lim     1
3  3  n   3 
  

3
 0  1  3 2
[since as n   ,    0 ]
n

 0  1 3

x x
x x x
lim  lim  lim
x  x  x 
x x x x  x x x  x  x
x x x x2 x4
1 1
 lim  1
x 
1 1  1 1 0  0
x x3

3x  1 3 1
lim  lim x  3 0  3
x  2 x  5 x 
2 5 20 2
x

x 22 2 1 2 2
x2 x x x x 00
lim  lim  lim  0
x  x 2  2 x  1 x 
1 2 x 2  1 2 x 
1 2 1  1 2 1 0  0
x
x x x
 6 3  13
6  t3 0 1 1
lim 3  lim t  
t  7t  3 t 
7 3 3 70 7
t

1 1
 2 2  3 5 3
 2 2  3 5 3
2  3 x  5 x 2
x x
lim 3  lim  x  
 lim x 
x  1  8x 2 x   1 8   x 1 2 8 
 x2   x 
1 1
 005 3
 5  3 3
5
    
 08   8  2

5x2  2 5 2 2 5 2 2
lim  lim x  lim x  50   5
x  x3 x  x3 x 
1  3 1  0
x
x

5x2  2 5 2 2 5 2 2
lim  lim x  lim x  50  5
x  x3 x  x3 x 
1 3 1 0
x
x

2 y 2
 1
2 y y y 0 1 1
lim  lim  lim  
y 
7  6 y2 y 
7  6 y2 y  7 06 6
2
6
y2 y

2 y 2
1
2 y y y 0 1 1
lim  lim  lim  
y 
7  6 y2 y 
7  6 y2 y  7 06 6
2
6
y2 y
  3 
lim
x 
 2

x  3  x  lim 
x 
 x2  3  x2 
 x 3 x
2
  lim
x 

 2
3 
  lim
 x 3  x

x  
x 

 1 3 2 1
 x 
 0 
 0
 1 0 1
 Continuity
A function f is said to be continuous on a closed interval [a,b] if the
following conditions are satisfied
1. f is continuous on (a,b)
2. f is continuous from the right at a i.e., lim f  x   f  a 
x a

3. f is continuous from the left at b i.e., lim f  x   f  b 

xb

Theorem:
 A polynomial function is continuous at every real number.
 A rational function q=f/g is continuous at every number except that
numbers c such that g(c)=0.
 A rational function has discontinuities at the points where the
denominator is zero.
 The function b x is continuous on its domain (-∞,∞).
 The function log b x is continuous on its domain (0,∞).
Theorem: If the functions f and g are continuous at c then
 f +g is continuous at c
 f -g is continuous at c
 f g is continuous at c
 f /g is continuous at c if g(c)≠0
Continuity on an interval: A function f is said to be continuous on a
closed interval [a,b] if the following conditions are satisfied
1. f is continuous on (a,b)
2. f is continuous from the right at a i.e., lim f  x   f  a 
xa

3. f is continuous from the left at b i.e., lim f  x   f  b 

xb
Question: If f  x   9  x 2 , sketch the graph of f and prove that f is

continuous in its domain [-3,3].

Solution: The domain of the given function is [-3,3]. We need to
investigate the continuity of f on the open interval (-3,3) and at the
two end points.
3

-3 -2 -1 1 2 3

If 3  c  3 then

lim f  x   lim 9  x 2  lim  9  x 2   9  c   f c 

2
x c x c x c

Hence f is continuous at c i.e. f is continuous on the open interval (-

3,3).
All that remains to check the continuity of the function at the two end
points.
lim f  x   lim 9  x 2  0  f  3 
x 3 x 3

lim f  x   lim 9  x 2  0  f  3 
x 3 x 3

Thus f is continuous on the closed interval [-3,3].

 9  x2
Question: If f  x   4 , sketch the graph of f and prove
3x  5 x 2  1
that f is continuous in its domain [-3,3].

Solution: Let g  x   9  x 2 and h  x   3 x 4  5 x 2  1.

From the above example g(x) is continuous on the closed interval [-

3,3].
Since h(x) is a polynomial function, it is continuous everywhere.
Moreover, h(c)≠0 for every number c in [-3,3].
Hence by theorem (a rational function q=f/g is continuous at every
number except that numbers c such that g(c)=0) the quotient

9  x2
f  x  4 is continuous on [-3,3].
3x  5 x 2  1
3

-3 -2 -1 1 2 3
 tan 1 x  ln x
Question: Where is the function f  x   continuous?
x2  4
Solution: The function will be continuous at all points where the
numerator and the denominator are both continuous and the
denominator is nonzero. Since tan 1 x is continuous everywhere and
ln x is continuous if x  0 , the numerator is continuous if x  0
(why?).
The denominator ( x 2  4 ) being a polynomial, is continuous
everywhere.
Therefore, the function f(x) will be continuous at all points where
x  0 and the denominator is nonzero.
Thus f(x) is continuous on the intervals (0,2)and (2,∞).
2

1 2 3 4

-1

-2

If 3  c  3 then
lim f  x   lim 9  x 2  lim  9  x 2   9  c   f c 
2
x c x c x c

Hence f is continuous at c i.e. f is continuous on the open interval (-

3,3).
All that remains to check the continuity of the function at the two end
points.
lim f  x   lim 9  x 2  0  f  3 
x 3 x 3

lim f  x   lim 9  x 2  0  f  3 
x 3 x 3
Thus f is continuous on the closed interval [-3,3]
1. Discuss the concept of limit. Find the limit (if it exists)

 1   1.8 x − 1.8− x 
i) lim   ii) lim  x 
x→ 0  1
 x →∞ 1.8 + 1.8− x
 1 − e5x   

5cos( x − 1), x > 1


=
iii) Let h( x) =a, x 1
 x + b, x <1

If h(x) is continuous everywhere, find the values of a and b.
2. Define continuity of a function at a point and on a closed interval. Suppose
that

− x 4 + 3, x ≤ 2
f ( x) =  2 and g ( x) = ( x + 1) / ( x 2 + x − 2)
 x + 9, x > 2

Are f and g continuous everywhere? Justify your conclusions.

3. Where are the following functions continuous?
 3 x 2 + sin x
 , x≠0
4 x 1
i) f ( x) =  ii)=
g ( x) + tan −1 ( x)
3 , (2 x − 1)
x=0
 4

9 − x2
4. If g ( x) = . Prove that g is continuous on [-3,3].
3x 4 + 5 x 2 + 1
5. Define differentiability of a function at a point and on a closed interval.
Check the differentiability of the function, f ( x)= x − 1 at x = 1 .

6. Determine the points, if any, at which the given function is discontinuous

(x − 9 x + 18 )
−1
i) f ( x) =x3 − 4 x 2 + 7 ii) f ( x) = 2

x −1  sin x
iii) f ( x) =
sin 2 x  x , x ≠ 0
iv) f ( x) = 
1 , x=0
 2

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7. Determine whether the given function is continuous on the indicated intervals
i) f ( x=
) x2 + 1 a) [-1, 4], conts b) [5, ∞), conts

1
ii) f ( x) = a) (0, 4], conts b) [1, 9], conts
x
iii) f ( x) = tan x a) [0, π], not conts b) [-π/2, π/2], not conts

x a) [-4, -3], conts b) (-∞, ∞), not conts

iv) f ( x) =
x +8
3

x
v) f ( x) = a) (-∞, ∞), not conts b) [π/2, 3π/2], not conts
2 + sec x

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