Cover slide
Limits
Limits
Limit is the value that a function "approaches"
as the input "approaches" some value.
❑ Limit of a function 𝑓 𝑥 is said to exist finitely as 𝑥 → 𝑎 when
lim − 𝑓 𝑥
𝑥→𝑎
= lim + 𝑓 𝑥 = 𝐿 (finite)s
Left hand
limit
Note:
𝑥→𝑎
Right hand
limit
lim − 𝑓 𝑥 = lim 𝑓 𝑎 − ℎ ; ℎ > 0
𝑥→𝑎
ℎ→0
lim + 𝑓 𝑥 = lim 𝑓 𝑎 + ℎ ; ℎ > 0
𝑥→𝑎
ℎ →0
Note:
❑ For polynomial functions, we can directly substitute the value of 𝑥.
❑ Also for trigonometric, logarithmic, exponential and
inverse trigonometric functions, we can directly
substitute the value of 𝑥 in their respective domain.
Remember:
lim 𝑎 𝑥 = 0 ; 0 < 𝑎 < 1
𝑥→∞
Example:
•
•
lim ln 𝑥 + 2
𝑥→0
lim𝜋 sin 2𝑥
𝑥→
2
•
lim1 sin−1 𝑥
𝑥→
•
2
lim 2 𝑥
𝑥→2
Let 𝑓 𝑥 = ቊ
cos 𝑥 , 𝑥 ≥ 0
, if lim 𝑓 𝑥 exists,
𝑥+𝑘, 𝑥 < 0
𝑥→0
then 𝑘 is equal to _______.
cos 𝑥 , 𝑥 ≥ 0
Given : 𝑓 𝑥 = ቊ
𝑥+𝑘, 𝑥<0
To find : lim 𝑓 𝑥
𝑥→0
𝑅. 𝐻. 𝐿. = lim+ 𝑓 𝑥 = lim+ cos 𝑥 = 1
𝑥→0
𝑥→0
𝐿. 𝐻. 𝐿. = lim− 𝑓 𝑥 = lim− 𝑘 + 𝑥 = 𝑘
𝑥→0
𝑥→0
Given that lim 𝑓 𝑥 exists
𝑥→0
⇒ 𝐿. 𝐻. 𝐿. = 𝑅. 𝐻. 𝐿.
⇒𝑘=1
Let 𝑓 𝑥 = ቊ
cos 𝑥 , 𝑥 ≥ 0
, if lim 𝑓 𝑥 exists,
𝑥+𝑘, 𝑥 < 0
𝑥→0
1
then 𝑘 is equal to _______.
2
𝑥2
−
The value of lim 9
𝑥→0
a
𝟎
b
𝟏
c
𝟗
d
Does not exist
is:
To find: lim
2
𝑥→0
−
9 𝑥2
2
lim+
𝑅. 𝐻. 𝐿 = 𝑥→0
∵ 𝑥 → 0+ , −
−
9 𝑥2
2
𝑥2
=0
→ −∞
2
𝐿. 𝐻. 𝐿 = lim− 9
𝑥→0
∵ 𝑥 → 0− , −
2
𝑥2
− 2
𝑥
=0
→ −∞
∴ 𝐿. 𝐻. 𝐿. = 𝑅. 𝐻. 𝐿. = 0
2
𝑥2
−
The value of lim 9
𝑥→0
a
𝟎
b
𝟏
c
𝟗
d
Does not exist
is:
The set of values of 𝑎 for which 𝑎2 − 𝑎 − 2 < 0 and lim 𝑎
𝑥→∞
a
−𝟏, 𝟎
b
𝟎, 𝟏
c
−𝟏, 𝟐
d
−𝟏, 𝟏
𝑥
= 0 is:
Given : 𝑎 2 − 𝑎 − 2 < 0 and lim 𝑎
𝑥→∞
⇒ 𝑎 + 1 𝑎 − 2 < 0 and 𝑎 < 1
⇒ −1 < 𝑎 < 2 and −1 < 𝑎 < 1
∴ 𝑎 ∈ −1, 1
𝑥
=0
The set of values of 𝑎 for which 𝑎2 − 𝑎 − 2 < 0 and lim 𝑎
𝑥→∞
a
−𝟏, 𝟎
b
𝟎, 𝟏
c
−𝟏, 𝟐
d
−𝟏, 𝟏
𝑥
= 0 is:
Algebra of Limits
Algebra of Limits
Let lim 𝑓(𝑥) = 𝑙, and lim 𝑔(𝑥) = 𝑚. If 𝑙 and 𝑚 are finite, then :
𝑥→𝑎
𝑥→𝑎
❑ lim {𝑓 𝑥 ± 𝑔 𝑥 } = 𝑙 ± 𝑚
𝑥→𝑎
❑ lim {𝑓 𝑥 ∙ 𝑔 𝑥 } = 𝑙 ∙ 𝑚
𝑥→𝑎
𝑓 𝑥
❑ lim 𝑔
𝑥→𝑎
𝑥
=
𝑙
𝑚
,𝑚 ≠ 0
❑ lim 𝑘𝑓 𝑥 = 𝑘 lim 𝑓 𝑥 = 𝑘𝑙 , 𝑘 is a constant
𝑥→𝑎
𝑥→𝑎
𝑥2 + 𝑥 − 6
𝑥→−1 𝑥 + 3
The value of lim
a
𝟑
b
−𝟑
c
𝟏
d
Limit does not exist
is:
To find: lim
𝑥2 + 𝑥 − 6
𝑥→−1
𝑥+3
We know that, lim
𝑓 𝑥
𝑥→𝑎 𝑔 𝑥
⇒ lim
𝑥→−1
𝑥2 + 𝑥 − 6
𝑥+3
=
lim 𝑓 𝑥
𝑥→𝑎
lim 𝑔 𝑥
𝑥→𝑎
lim 𝑥2 + 𝑥 − 6
𝑥→−1
lim 𝑥 + 3
𝑥→−1
=
=
1−1 −6
−1 + 3
= −3
lim 𝑔 𝑥 ≠ 0
; 𝑥→𝑎
𝑥2 + 𝑥 − 6
𝑥→−1 𝑥 + 3
The value of lim
a
𝟑
b
−𝟑
c
𝟏
d
Limit does not exist
is:
Indeterminate Forms
Indeterminate Forms
Whose value can’t be directly determined by substituting value of 𝑥.
We have 7 indeterminate forms :
Note:
Forms like :
❑ ∞×∞→∞
❑ ∞+∞→∞
❑
❑
❑
𝑎
= 0 (if 𝑎 is finite)
∞
∞
→0
❑ ∞∞ → ∞
❑ exact 0
→0+
=0
−
❑ (exact 0)→0 = not defined
→ ±∞
exact 0
→0
❑ → 0+
are not indeterminate forms.
=0
∞
=0
Methods to Solve Indeterminate Forms
❑ Factorization/Rationalization
❑ Using Substitution
𝑥+1−1
𝑥
𝑥→0
The value of lim
a
b
𝟐
𝟑
−
𝟏
𝟐
c
𝟏
d
𝟏
𝟐
is:
To find: lim
= lim
𝑥→0
= lim
𝑥→0 𝑥
= lim
𝑥→0
=
1
2
𝑥+1−1
𝑥→0
𝑥
𝑥 + 1 −1
×
𝑥
𝑥+1−1
𝑥+1+1
1
𝑥 + 1 +1
0
0
form
𝑥+1+1
𝑥+1+1
[Using rationalisation]
𝑥+1−1
𝑥
𝑥→0
The value of lim
a
b
𝟐
𝟑
−
𝟏
𝟐
c
𝟏
d
𝟏
𝟐
is:
The value of
a
𝟐
b
𝟑
c
𝟓
d
𝟕
𝟑
𝟐
𝟒
𝟒
𝑥3 − 𝑥2 ln 𝑥 + ln 𝑥 − 1
lim
𝑥2 − 1
𝑥→1
is:
To find: lim
𝑥3 − 𝑥2 ln 𝑥 + ln 𝑥 − 1
0
𝑥2 − 1
0
𝑥→1
= lim
𝑥→1
= lim
𝑥→1
= lim
𝑥→1
=
3
2
𝑥3 −1− 𝑥2 −1 ln 𝑥
𝑥2 −1
𝑥−1 𝑥2 +1+𝑥 − 𝑥−1 𝑥 +1 ln 𝑥
𝑥−1 𝑥+1
𝑥2 + 1 + 𝑥 − 𝑥 + 1 ln 𝑥
𝑥+1
form
The value of
a
𝟐
b
𝟑
c
𝟓
d
𝟕
𝟑
𝟐
𝟒
𝟒
𝑥3 − 𝑥2 ln 𝑥 + ln 𝑥 − 1
lim
𝑥2 − 1
𝑥→1
is:
Standard Limits
Standard Limits
sin 𝑥
=1
❑ lim
𝑥→0 𝑥
sin−1 𝑥
=1
❑ lim
𝑥→0
𝑥
tan 𝑥
=1
𝑥→0 𝑥
tan −1 𝑥
=1
❑ lim
𝑥→0
𝑥
❑ lim
Note:
sin 𝑎𝑥
tan 𝑎𝑥 𝑎
= lim
=
𝑥→0 𝑏𝑥
𝑥→0 𝑏𝑥
𝑏
❑ lim
sin−1 𝑎𝑥
tan−1 𝑎𝑥 𝑎
❑ lim
= lim
=
𝑥→0
𝑥→0
𝑏𝑥
𝑏𝑥
𝑏
Standard Limits
𝑥 𝑛 − 𝑎𝑛
❑ lim
= 𝑛𝑎 𝑛−1
𝑥→𝑎 𝑥 − 𝑎
𝑎𝑥 − 1
❑ lim
= ln 𝑎
𝑥→0
𝑥
e𝑥 − 1
❑ lim
=1
𝑥→0
𝑥
❑ lim
ln 1 + 𝑥
=1
𝑥→0
𝑥
Note:
ⅇ 𝑎𝑥 − 1 𝑎
❑ lim
=
𝑥→0
𝑏𝑥
𝑏
𝑎 𝑘𝑥 − 1 𝑘
❑ lim
= ln 𝑎
𝑥→0
𝑚𝑥
𝑚
ln 1 + 𝑎𝑥
𝑎
=
𝑥→0
𝑏𝑥
𝑏
❑ lim
Statement 𝐼 ∶ lim
𝑥→0
1 − cos 2𝑥
does not exist
𝑥
Statement 𝐼𝐼 ∶ 𝑓 𝑥 =
1−cos 2𝑥
𝑥
is not defined at 𝑥 = 0
a
Both 𝑰 and 𝑰𝑰 are individually true and
𝑰𝑰 is the correct explanation of 𝑰
b
Both 𝑰 and 𝑰𝑰 are individually true and
𝑰𝑰 is not the correct explanation of 𝑰
c
𝑰 is true but 𝑰𝑰 is false
d
𝑰 is false but 𝑰𝑰 is true
lim
Statement 𝐼 ∶ 𝑥→0
1 − cos 2𝑥
does not exist
𝑥
Let 𝐿 = lim
1 − cos 2𝑥
𝐿𝐻𝐿 = lim−
2 sin 𝑥
𝑥
𝑥→0
𝑥→0
= lim
𝑥
− 2 sin 0−ℎ
ℎ→0
= lim −
ℎ→0
=− 2
0−ℎ
2 sin ℎ
ℎ
= lim
𝑥→0
2 sin 𝑥
𝑥
lim+
𝑅𝐻𝐿 = 𝑥→0
= lim
2 sin 𝑥
𝑥
− 2 sin 0+ℎ
0+ℎ
ℎ→0
= lim
ℎ→0
2 sin ℎ
ℎ
= 2
∴ 𝐿𝐻𝐿 ≠ 𝑅𝐻𝐿
Hence, the limit for the function does not exist.
Statement 𝐼𝐼 ∶ 𝑓 𝑥 =
1−cos 2𝑥
𝑥
is not defined at 𝑥 = 0
Statement 𝐼𝐼 is also true, but it is not the correct explanation of statement 𝐼.
As for limit to exist, it is not necessary that function be defined at that point.
Statement 𝐼 ∶ lim
𝑥→0
1 − cos 2𝑥
does not exist
𝑥
Statement 𝐼𝐼 ∶ 𝑓 𝑥 =
1−cos 2𝑥
𝑥
is not defined at 𝑥 = 0
a
Both 𝑰 and 𝑰𝑰 are individually true and
𝑰𝑰 is the correct explanation of 𝑰
b
Both 𝑰 and 𝑰𝑰 are individually true and
𝑰𝑰 is not the correct explanation of 𝑰
c
𝑰 is true but 𝑰𝑰 is false
d
𝑰 is false but 𝑰𝑰 is true
The value of
a
𝟏
𝟐
b
𝟐
c
𝟐
d
1 − cos 𝑥2
lim
𝑥→0 1 − cos 𝑥
Does not exist
is:
1 − cos 𝑥2
To find: lim
𝑥→0 1 − cos 𝑥
= lim
𝑥→0
𝑥2
2
𝑥
2
2 sin
2
2 sin 2
𝑥2
2
𝑥
2
𝑥→0 2 sin 2
= lim
= lim
𝑥→0
2 sin
𝑥2
2
𝑥
2 sin2
2
sin
∵ sin
=
2
2
𝑥2
2
is always positive
= 2
∵ lim
𝑥→0
sin 𝑥
𝑥
=1
The value of
a
𝟏
𝟐
b
𝟐
c
𝟐
d
1 − cos 𝑥2
lim
𝑥→0 1 − cos 𝑥
Does not exist
is:
tan 𝑥−1 +1− 𝑥
𝑥−1 1−𝑥
𝑥→1
The value of lim+
is:
(where ⋅ denotes greatest integer function)
a
−𝟏
b
𝟎
c
𝟏
d
Does not exist
tan 𝑥−1 +1− 𝑥
𝑥−1 1−𝑥
𝑥→1+
To find: lim
= lim
tan 1+ℎ −1 +1− 1+ℎ
= lim
tan ℎ+1−1
1+ℎ−1 1−1−ℎ
ℎ→0
ℎ→0
= lim −
ℎ→0
= −1 × 1
= −1
−ℎ
tan ℎ
ℎ
tan 𝑥−1 +1− 𝑥
𝑥−1 1−𝑥
𝑥→1
The value of lim+
is:
(where ⋅ denotes greatest integer function)
a
−𝟏
b
𝟎
c
𝟏
d
Does not exist
THANK YOU
P US H YO URS E L F
TO B E GRE A T