Indeterminate Form
MATH 1013 Calculus I (Lecture 16)
Yichao Zhu
Department of Mathematics, HKUST
Goals
Use the L’Hospital’s Rule to determine the limit of
Indeterminate Form.
We have experienced the indeterminate form of type 0/0 or ∞/∞
such as
sin x
ln x
ln x
lim
, lim
, lim
,...
x→0 x
x→1 x − 1 x→∞ x − 1
Some of the limits actually exist.
Question
Can you think of other types of indeterminate forms?
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
L’Hospital’s Rule
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
Example 1
L’Hospital’s Rule
The L’Hospital’s Rule gives a general approach to determine the
limit of the indeterminate forms.
Remarks
I
I
L’Hospital Rule is also valid for one-sided limits and for limits
at ±∞
Example
Check the conditions of L’Hospital Rule and if met use it to
calculate
ln x
1. lim
x→1 x − 1
sin x
2. lim
−
x→π 1 − cos x
ln x
3. lim 3 √
x→∞
x
tan x − x
4. lim
x→0
x3
x
e
5. lim 2
x→∞ x
Verify the conditions regarding the limits of f and g
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
Example 1
Example 1
Solution - Part 1
Solution - Part 2
1. First check lim ln x = 0 and lim x − 1 = 0, applicable.
x→1
Then by L’Hospital’s Rule,
x→1
4. First check lim tan x − x = 0 and lim x 3 = 0, applicable.
x→0
tan x − x
sec2 x − 1
sin2 x
=
lim
=
lim
x→0
x→0
x→0 3x 2 cos2 x
x3
3x 2
2
1
sin x
1
1
· lim
lim
=
=
x→0 cos2 x
3 x→0 x
3
lim
(ln x)0
ln x
1
= lim
= lim = 1.
0
x→1 (x − 1)
x→1 x − 1
x→1 x
lim
2. First check lim sin x = 0 and lim (1 − cos x) = 2, not
x→π −
x→π −
applicable. In fact, by direct substitution
x→0
5. First check lim e x = ∞ and lim x 2 = ∞, applicable
x→∞
sin x
0
lim
= = 0.
−
2
x→π 1 − cos x
3. First check lim ln x = ∞ and limx→∞ x 1/3 = ∞, applicable.
x→∞
lim
ex
ex
=
lim
(still an undeterminate form)
x→∞ x 2
x→∞ 2x
lim
since lim e x = ∞ and lim 2x = ∞, by L’Hospital’s Rule
x→∞
ln x
1/x
√ = lim −2/3 = lim 3x −1/3 = 0.
x→∞ x
x
/3 x→∞
MATH 1013 Calculus I (Lecture 16)
Other Types of Indeterminate Form and Example 2
x→∞
ex
ex
= lim
= ∞.
x→∞ 2x
x→∞ 2
lim
x→∞ 3
Yichao Zhu Department of Mathematics, HKUST
x→∞
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
Example 2
Indeterminate Product
If limx→a f (x) = 0 and limx→a g (x) = ±∞, then limx→a f (x)g (x)
becomes an indeterminate form of type 0 · ∞.
To use L’Hospital’s Rule, we can change it to 0/0 or ∞/∞ by
f
g
fg =
or fg =
.
1/g
1/f
Indeterminate Difference
If limx→a f (x) = ∞ and limx→a g (x) = ∞, then
limx→a f (x) − g (x) becomes an indeterminate form of type
∞ − ∞. To use L’Hospital’s Rule, we convert the difference into a
quotient (e.g. by using a common denominator or rationalisation).
x→0
1. As x → 0+ , ln x → −∞ and x → 0, but 1/x → ∞. Therefore,
lim+ x ln x = lim+
x→0
x→0
ln x
x −1
= lim+
= − lim+ x = 0.
1/x
x→0 −x −2
x→0
2. As x → (π/2)− , sec x → ∞ and tan x → ∞. It is of ∞ − ∞
type
lim
x→(π/2)−
sec x − tan x =
=
Example
Evaluate
1. lim+ x ln x
Solution
2.
Yichao Zhu Department of Mathematics, HKUST
lim
x→(π/2)−
lim
x→(π/2)−
lim
x→(π/2)−
1 − sin x
(of type 0/0)
cos x
− cos x
= 0.
− sin x
(sec x − tan x)
MATH 1013 Calculus I (Lecture 16)
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
Other Types of Indeterminate Form and Example 3
Indeterminate Power
Several indeterminate forms also arise from the limit lim [f (x)]g (x)
x→a
1. lim f (x) = 0 and lim g (x) = 0, type 00
x→a
x→a
2. lim f (x) = ∞ and lim g (x) = 0, type ∞0
x→a
x→a
3. lim f (x) = 1 and lim g (x) = ±∞, type 1±∞
x→a
x→a
One can transform the above forms into indeterminate products by
taking logarithm
Let y = [f (x)]g (x) , then ln y = g (x) ln f (x).
Example 2
Solution
1. Let y = (1 + sin 4x)cot x , then ln y = cot x ln(1 + sin 4x). By
L’Hospital’s Rule,
lim+ ln y = lim+
x→0
x→0
4 cos 4x
ln(1 + sin 4x)
= lim+
= 4.
2
tan x
x→0 sec x(1 + sin 4x)
Thus lim+ (1 + sin 4x)cot x = e limx→0+ ln y = e 4 .
x→0
2. Let y = x x , then ln y = x ln x. From one of the previous
examples, we know
Thus the original limit becomes
lim x ln x = 0.
lim [f (x)]g (x) = e limx→a g (x) ln[f (x)] .
Example
x→a
1. lim+ (1 + sin 4x)cot x
x→0
2. lim+ x x
Yichao Zhu Department of Mathematics, HKUST
x→0
MATH 1013 Calculus I (Lecture 16)
Homework
Web of Work
I
Worksheet 6 (deadline: 15-NOV-2014)
I
Worksheet 7 (deadline: 23-NOV-2014)
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)
x→0+
Thus
lim x x = e limx→0+ x ln x = 1.
x→0+
Yichao Zhu Department of Mathematics, HKUST
MATH 1013 Calculus I (Lecture 16)

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