Welcome to
Limits
Table of contents
Session 01 03
Limits 04
Indeterminate Forms 13
Algebra of limits 15
Methods to solve 16
Rationalisation 20
Session 02 23
Sandwich theorem 29
Important deductions 33
Standard Limits 37
Session 1
Introduction to Limits and
methods to solve
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Key Takeaways
Limits:
• If 𝑥 → 𝑎 gives 𝑓 𝑥 → 𝑚 then it is represented by lim 𝑓 𝑥 = 𝑚
!→#
• Here 𝑚 is the limiting value of 𝑓 𝑥 at 𝑥 = 𝑎
𝑥 → 𝑎 means 𝑥 tends to 𝑎
Key Takeaways
𝑌
As 𝑥 tends to 3, 𝑓(𝑥)
tends to 3.99
lim 𝑓 𝑥 = 3.99
!→$
(2.99,3.99)
(3.01,3.99)
𝑋
Key Takeaways
Limits:
• If 𝑥 → 𝑎 gives 𝑓 𝑥 → 𝑚 then it is represented by lim 𝑓 𝑥 = 𝑚
!→#
• Here 𝑚 is the limiting value of 𝑓 𝑥 at 𝑥 = 𝑎
• 𝑥 → 𝑎( 𝑥 ≠ 𝑎) means that 𝑥 approaches to 𝑎
• For RHL 𝑥 = 𝑎 + ℎ
𝑥 =𝑎−ℎ 𝑥 =𝑎+ℎ
For LHL 𝑥 = 𝑎 − ℎ
𝑎
• ℎ is infinitely small +𝑣𝑒 quantity
Key Takeaways
Note:
• lim 𝑓 𝑥 = lim 𝑓 𝑎 − ℎ
!→# ! %→&
• lim 𝑓 𝑥 = lim 𝑓 𝑎 + ℎ 𝑎−ℎ 𝑎 𝑎+ℎ
!→# " %→&
lim 𝑓 𝑥 = lim" 𝑓 𝑥 = 𝑙(:inite)
!→# ! !→#
Key Takeaways
10
𝑓(𝑥)
1 𝑋
−5 −1 1 5 10 15 20
• Limit does not exist at 𝑥 = 5 and 𝑥 = 24 because L.H.L ≠ R.H.L
• Limit exists at 𝑥 = 13 & 18 because L.H.L = R.H.L
𝑥 + 1, 𝑥 ≤ 1
If 𝑓 𝑥 = % , then evaluate
2𝑥 − 3, 𝑥 > 1
a) lim 𝑓 𝑥 b) lim 𝑓 𝑥
0→2 0→3
𝑥 + 1, 𝑥 ≤ 1
If 𝑓 𝑥 = B evaluate lim 𝑓 𝑥
2𝑥 − 3, 𝑥 > 1 !→'
a) lim 𝑓 𝑥
!→'
lim 𝑓 𝑥 = lim 1 − ℎ + 1 = 2
!→'! %→&
2𝑥 − 3
lim 𝑓 𝑥 = lim 2 1 + ℎ − 3 = − 1 𝑥+1
!→'" %→&
Since, L.H.L ≠ R.H.L 2
∴ lim 𝑓(𝑥) = does not exist
!→' 1
−1
b) lim 𝑓 𝑥
!→(
lim 𝑓 𝑥 = lim 2 4 − ℎ − 3 = 5
!→( ! %→&
lim 𝑓 𝑥 = lim 2 4 + ℎ − 3 = 5
!→( " %→&
Since, L.H.L = R.H.L
∴ lim 𝑓(𝑥) = 5
!→(
Evaluate lim −1 0
0→22
!
lim −1
!→''
L.H.L 𝑥 = 11 − ℎ R.H.L 𝑥 = 11 + ℎ
'')% '& ''*% ''
lim −1 = −1 = +1 lim −1 = −1 = −1
%→& %→&
Limit does not exist
0
Evaluate lim
0→4 0 50+
L.H.L R.H.L
𝑥 =0−ℎ 𝑥 =0+ℎ
% %
lim − #
lim
%→& )% * )% %→& % * % #
)% %
lim lim
%→& %*% # %→& %*% #
)% %
lim lim
%→& % '*% %→& % '*%
)' ' ' '
lim = − '*& = −1 lim = '*& = 1
%→& '*% %→& '*%
L.H.L ≠ R.H.L
Limit does not exist
Key Takeaways
Indeterminate forms:
• Whose value can′t be directly determined by
substituting value of 𝑥
• We have 7 indeterminate forms :
& ,
,
& ,
, ∞ − ∞, 0×∞, 1, , 0& , ∞&
Which of the following are indeterminate forms:
062 262 4
1. lim 0+62 = 262 = 4 Yes
0→2
2. lim 𝑥8 − 1 − 𝑥 = −∞ 8 − 1 − −∞ → ∞ No
0→67
Key Takeaways
Algebra of limits:
Let lim 𝑓(𝑥) = 𝑙, and lim 𝑔(𝑥) = 𝑚. If 𝑙 and 𝑚 are finite, then:
!→# !→#
• lim {𝑓 𝑥 ± 𝑔 𝑥 } = 𝑙 ± 𝑚 • lim 𝑘𝑓 𝑥 = 𝑘 lim 𝑓 𝑥 = 𝑘𝑙 , 𝑘 is a constant
!→# !→# !→#
Ex: lim 𝑥 + sin 𝑥 = 2 + sin 2 Ex: lim& 2 tan 𝑥 = 2 3
!→-
!→
$
• lim {𝑓 𝑥 . 𝑔 𝑥 } = 𝑙. 𝑚
!→#
$
$ ./0
Ex: lim$ 𝑥. cos 𝑥 = (
%
!→
%
1 ! 3
• lim 2 !
= 4 ,𝑚 ≠ 0
!→#
! # *'
Ex: lim =1
!→' !*'
Key Takeaways
Methods to solve the limit:
• Questions which do not have [],| |,{ }, defined function
need not be solved by comparing LHL,RHL
• So we just solve them as it is using following methods
(A) Factorization (E) Use of B.T
(B) Rationalization (F) Sandwich Theorem
(C) Double Rationalization (G) Trigo
(D) limit 𝑛 → ∞ (H) Expansion
0+6905:
Evaluate: lim
0→8 0+6;058
! # )5!*6 ()'&*6 &
lim = = form
!→- ! # )$!*- ()6*- &
(!)-)(!)$)
lim (!)-))(!)')
!→-
-)$
=
-)'
= −1
0;60+ <=> 05<=> 062
Evaluate: lim 0+62
0→2
! $ )' ) ! # )' 9/: !
= lim
!→' (! # )')
!)' ! # *!*') !*' 9/: !
= lim (!)')(!*')
!→'
! # *!*') !*' 9/: !
= lim (!*')
!→'
3
=
2
0+
Evaluate : lim
0→4 250+ 6 260+
!# '*! # * ')! #
= lim '*! # ) ')! #
!→&
!# '*! # * ')! #
= lim -! #
!→&
'*'
=
-
=1
Key Takeaways
Rationalization:
When given expression is in the form of 𝑎 − 𝑏 or 𝑎 −𝑏
form we do rationalization.
05368
Evaluate : lim
0→4 0
( !*()-)( !*(*-)
= lim !( !*(*-)
!→&
!*()(
= lim !( !*(*-)
!→&
'
= lim ( !*(*-)
!→&
'
=
(
;6 ?052
Evaluate : lim
0→2 96 83052
$) <!*'
lim
!→' 5) -(!*'
$) <!*' $* <!*' (5* -(!*')
= lim
!→' (5) -(!*')(5* -(!*') $* <!*'
(=) <!*' )(5* -(!*')
= lim
!→' (-5) -(!*' )($* <!*')
<×'&
=
-(×6
5
==
Session 2
Sandwich theorem and some
standard limits
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Key Takeaways
Basics:
,
• Less than 1 =0
,
• Greater than 1 →∞
,
• Exactly 1 =1
A A
lim 3@ + 4@ B is equal to lim 3@ + 4@ + 5@ + 6@ B is equal to
@→7 @→7
'/? '/?
$ ? ( ? $ ? ( ? 5 ?
= lim 4 + = lim 6 6
+ 6
+ 6
+1
?→, ( ( ?→,
&
=4 0+1 & =6 0+0+0+1
= 4×1 = 6×1
=4 =6
A0+62;052 A0+62;05: 8059
Evaluate: lim Evaluate: lim Evaluate: lim
0→7 90+6805: 0→7 8059 0→7 A0+62;05:
C! # )'$!*' C! # )'$!*6 -!*5
lim lim lim
!→, 5! # )-!*6 !→, -!*5 !→, C! # )'$!*6
'$ ' '$ ) *
!# C ) * # !# C ) * # ! -*
(
= lim (
#
(
) = lim (
*
(
= lim '$ )
!→, !# 5 ) * # !→, ! -* !→, !# C ) * #
( ( ( ( (
C -*&
=5 = 7×∞ = →0
,
DEFG9 ./HIJK.KHLM NK::HO PQG99HO
lim = ./HIJK.KHLM lim PQG99HO
→∞ lim →0
!→, DEFG9 !→, !→, NK::HO
If 𝑓: 𝑅 → 𝑅 is given by 𝑓 𝑥 = 𝑥 + 1, then the value of
1 5 10 5 𝑛−1
lim 𝑓 0 +𝑓 +𝑓 + ⋯+ 𝑓
?→, 𝑛 𝑛 𝑛 𝑛
C $
A - B -
5 '
C - D -
If 𝑓: 𝑅 → 𝑅 is given by 𝑓 𝑥 = 𝑥 + 1, then the value of
$ % $& % !'$
lim 𝑓 0 +𝑓 +𝑓 + ⋯+ 𝑓
!→# ! ! ! !
5 '& 5 ?)'
𝑓 0 +𝑓 ?
+𝑓 ?
+ ⋯+ 𝑓 ?
5 '& 5 ?)'
⇒1+ 1+ + 1+ + ⋯+ 1 +
? ? ?
5 ? ?)' -?*5?)5 C?)5
⇒𝑛+? -
= -
= -
' C?)5 C
⇒ lim =
?→, ? - -
Key Takeaways
Sandwich Theorem:
Let 𝑓, 𝑔, ℎ are 3 functions in small open interval,
such that 𝑓 𝑥 ≤ 𝑔 𝑥 ≤ ℎ 𝑥 and
𝑌
lim 𝑓 𝑥 = 𝑙 lim ℎ 𝑥 = 𝑙
!→# !→# ℎ(𝑥)
𝑙
Since 𝑓 𝑥 ≤ 𝑔 𝑥 ≤ ℎ 𝑥 𝑔(𝑥)
⇒ lim 𝑓 𝑥 ≤ lim 𝑔 𝑥 ≤ lim ℎ 𝑥 𝑓(𝑥)
!→# !→# !→#
⇒ 𝑙 ≤ lim 𝑔 𝑥 ≤ 𝑙 𝑎 𝑋
!→#
So, lim 𝑔 𝑥 = 𝑙
!→#
Proof:
Chord 𝐴𝐵 < Arc 𝐴𝐵 < 𝐴𝐶 + 𝐵𝐶
R%STU VW VTX VW VR*WR
⇒ < < 𝐵
- - -
g < 𝐵𝐶
⇒ 𝐵𝐸 < BD 𝜃𝐸 𝐷 𝐶
ND Z
N[ WR
O
⇒ YN < \W < \W
⇒ sin 𝜃 < 𝜃 < tan 𝜃 𝐴
0KL ! 0KL ] MGL ] MGL !
!
<1 < 1 < >1
] ] !
FGH 0 244 FGH 0 2440
Evaluate: lim Evaluate:lim +
0→4 0 0→4 0 IJH 0
lim less than 1 = 0 0KL !
<1
MGL !
>1⇒
!
<1
!→& ! ! MGL !
'&& 0KL ! '&&!
= !
< 100 = MGL !
< 100
= 99.99 = 99.99
= 99 = 99
⇒ 99 + 99 = 198
Key Takeaways
Proof:
0KL ] MGL ]
<1<
] ] 𝐵
When 𝜃 → 0
⇒ sin 𝜃 ≅ 𝜃 ≅ tan 𝜃 𝜃 𝐸 𝐷 𝐶
O
0KL ] MGL ]
⇒ lim = 1 & lim =1
]→& ] ]→& ]
𝐴
Remember: 𝜃 → 0
sin 𝜃 ≅ tan 𝜃 ≅ sin)' 𝜃 ≅ tan)' 𝜃 = 𝜃
Important Deductions:
0KL !
lim !
= 1 will be valid under 2 conditions
!→&
Same quantity should be there in denominator
and for argument/angle in numerator &
This quantity should be approaching zero
0KL 0GQH
(0GQH)
0KL !
lim !
= 1 only
!→&
0KL(&)
(&)
Similarly,
sin 𝑥 𝑥 tan 𝑥 𝑥
lim = lim = lim = lim =1
!→& 𝑥 !→& sin 𝑥 !→& 𝑥 !→& tan 𝑥
Evaluate the following :
FGH K0 FGH K0
a) lim b) lim
0→4 0 0→4 IJH L0
FGH K0 FGH K0
a) lim = 𝑎 lim =𝑎⋅1=𝑎
0→4 0 0→4 K0
FGH K0 K FGH K0 L0 K K
b) lim IJH L0 = lim
L 0→4 K0 IJH L0
=L⋅1⋅1=L
0→4
FGH M60
Evaluate : lim
0→M M(M60)
Solution:
If 𝑥 → 𝜋 then 𝜋 − 𝑥 → 0
Let 𝑡 = 𝜋 − 𝑥
Now,
0KL(^)!) 0KL _
lim = lim
!→^ ^ ^)! _→& ^⋅_
' 0KL _ '
= lim =
^ _→& _ ^
FGH; 80 IJH+ ;0
Evaluate : lim
0→4 0 FGHa 30
Solution:
As 𝑥 → 0 sin 𝑥 , tan 𝑥 ≅ 𝑥
So,
0KL$ -! MGL# $! -! $ $! # =
lim ! 0KL% (!
= = $-
!→& ! (! %
Standard Limits
𝑥 ? − 𝑎?
lim = 𝑛 ⋅ 𝑎?)'
!→# 𝑥 − 𝑎
Proof:
We know that
𝑥 ? − 𝑎? = 𝑥 − 𝑎 𝑥 ?)' + 𝑎 ⋅ 𝑥 ?)- + ⋯ + 𝑥 ⋅ 𝑎?)- + 𝑎?)'
! + )# +
⇒ !)#
= 𝑥 ?)' + 𝑎 ⋅ 𝑥 ?)- + ⋯ + 𝑥 ⋅ 𝑎?)- + 𝑎?)'
! + )# +
∴ lim = lim 𝑥 ?)' + 𝑎 ⋅ 𝑥 ?)- + ⋯ + 𝑥 ⋅ 𝑎?)- + 𝑎?)'
!→# !)# !→#
! + )# +
∴ lim = 𝑎?)' + 𝑎?)' + ⋯ 𝑛 times
!→# !)#
= 𝑛 ⋅ 𝑎?)'
𝑥 ? − 𝑎?
lim = 𝑛 ⋅ 𝑎?)'
!→# 𝑥 − 𝑎
𝑥 4 − 𝑎4 𝑚 4)?
lim = ⋅𝑎
!→# 𝑥 ? − 𝑎? 𝑛
𝑎! − 1
lim = log b 𝑎
!→& 𝑥
𝑒! − 1
lim =1
!→& 𝑥
ln 1 + 𝑥
lim =1
!→& 𝑥
<H 30+53052
Evaluate lim
0→4 0
Solution:
ln 4𝑥 - + 4𝑥 + 1 ln 1 + 2𝑥 -
lim = lim
!→& 𝑥 !→& 𝑥
2 ln(1 + 2𝑥) ln(1 + 2𝑥)
= lim = 4 lim
!→& 𝑥 !→& 2𝑥
=4⋅1=4
0c69c
If lim = 500, then the value of 𝑘 is
0→9 069
Solution:
! , )5,
lim = 𝑘 ⋅ 5d)' = 500
!→5 !)5
⇒ 𝑘 ⋅ 5d)' = 4 ⋅ 5$
⇒𝑘=4
8e6N e
Evaluate : lim
0→4 0
Solution:
We know that
# ( )' b ( )'
lim !
= log b 𝑎 and lim !
=1
!→& !→&
-( )b ( -( )' ) b ( )'
∴ lim = lim
!→& ! !→& !
-( )' b ( )'
= lim !
− lim !
!→& !→&
-
= log b 2 − 1 = log b
b
;e62
Evaluate : lim
0→4 25062
Solution:
$( )' $( )' '*!*'
lim '*!)'
= lim '*!)'
⋅ '*!*'
!→& !→&
$( )' '*!*'
= lim
!→& !
$( )'
= lim !
⋅ lim 1+𝑥+1
!→& !→&
= log b 3 ⋅ 2 = log b 9
Evaluate : lim 𝑥 8 + 3𝑥 − 𝑥
0→7
Solution:
lim 𝑥 - + 3𝑥 − 𝑥
!→,
! # *$!)! ! # *$!*!
= lim
!→, ! # *$!*!
$!
= lim
!→, ! '* *'
$
(
$
= lim
!→, $
'* *'
(
$
=-
