Indeterminate Forms

and L’Hopital’s Rule

Section 4.4

Prepared by:
Prof. Wafik Lotfallah
Prof. Maged G. Iskander
Dr. Kamal Tawfik
Lecture 20 Objectives
 Decide if a given form is indeterminate.
 Know the conditions under which L’Hopitals’ Rule can
be applied.
 Apply L’Hopital’s Rule to find limits of the form:
 0/0
 ∞/∞.
 Apply the natural logarithm and L’Hopital’s Rule to
find limits of the form: 00 , ∞0 , 1∞
L’Hopital`s Rule and Indeterminate Forms
Back in the chapter on Limits we saw methods for dealing
with the following limits
sin 𝑥 4𝑥 2 − 3𝑥
lim and lim
𝑥→0 𝑥 𝑥→∞ 1 − 3𝑥 2
In the first limit if we plugged in 𝑥 = 0 we would get 𝟎/𝟎
and in the second limit if we “plugged” in infinity we would
get ∞/∞ .

Both of these are called indeterminate forms. In both of

these cases there are rules that tell us what should happen.
 𝟎 ∞
Case 1: Indeterminate Forms &
𝟎 ∞
L’Hopital’s Rule

So, L’Hopital’s Rule tells us that if we have an indeterminate

form 0/0 or ∞/∞ all we need to do is differentiate the
numerator and differentiate the denominator and then take
the limit.
Example 1
ln 𝑥
Find lim .
𝑥→1 𝑥−1
Solution
Since,
Example 2
𝑒𝑥
Calculate lim 2 .
𝑥→∞ 𝑥
Solution
Since, lim 𝑒 𝑥 = ∞ and lim 𝑥 2 = ∞ , so the limit is an
𝑥→∞ 𝑥→∞
indeterminate form of type ∞Τ∞, and l’Hopital’s Rule gives

Again, lim 𝑒 𝑥 = ∞ and lim 2𝑥 = ∞ , the limit on the right side

𝑥→∞ 𝑥→∞
is also indeterminate, but a second application of l’Hopital’s
Rule gives
Example 3
tan 𝑥 − 𝑥
Calculate lim .
𝑥→0 𝑥3
Solution
0
Clearly the limit has the form , then
0
tan 𝑥 − 𝑥 𝐻 sec 2 𝑥 − 1
lim ֜ = lim
𝑥→0 𝑥3 𝑥→0 3𝑥 2
Since the limit on the right side is still indeterminate of type
0/0 we apply l’Hopital’s Rule again:
sec 2 𝑥 − 1 𝐻 2 sec 2 𝑥 tan 𝑥
lim 2
֜ = lim
𝑥→0 3𝑥 𝑥→0 6𝑥
1 2
tan 𝑥
= lim sec 𝑥 lim
3 𝑥→0 𝑥→0 𝑥
1
= 1 1 = 1Τ3 .
3
Case 2: Indeterminate Forms 𝟎. (±∞)
The indeterminate form 𝟎. (±∞) occurs if we have the limit
of product of two functions, i.e., lim 𝑓 𝑥 𝑔(𝑥)
𝑥→𝑎
such that lim 𝑓 𝑥 = 0 𝑎𝑛𝑑 lim 𝑔(𝑥) = ∞ (0𝑟 − ∞) .
𝑥→𝑎 𝑥→𝑎

In this case we always write a product of functions as a

quotient by doing one of the following:
𝑓 𝑔
𝑓𝑔 = 𝑜𝑟 𝑓𝑔 =
1Τ𝑔 1Τ𝑓

This converts the given limit into an indeterminate form of

𝟎
type or ∞Τ∞ so that we can use l’Hospital’s Rule.
𝟎
Example 4
Evaluate lim+ 𝑥 ln 𝑥 .
𝑥→0
Solution
Clearly the limit has the form 0. (−∞), since
lim+ 𝑥 = 0 and lim+ ln 𝑥 = −∞
𝑥→0 𝑥→0
ln 𝑥
So, we write 𝑥 ln 𝑥 as .
1Τ𝑥
Then, l’Hopital’s Rule gives
ln 𝑥 1Τ𝑥
lim+ 𝑥 ln 𝑥 = lim+ Τ = lim+ Τ 2 = lim+ (−𝑥) = 0 .
𝑥→0 𝑥→0 1 𝑥 𝑥→0 −1 𝑥 𝑥→0
Note:
In solving Example 4 another possible option would have been
to write

𝟎
This gives an indeterminate form of the type , but if we
𝟎
apply l’Hopital’s Rule we get a more complicated expression
than the one we started with.

In general, when we rewrite an indeterminate product, we try

to choose the option that leads to the simpler limit.
Case 3: Indeterminate Form ∞ − ∞
If lim 𝑓 𝑥 = ∞ and lim 𝑔 𝑥 = ∞ , then the limit
𝑥→𝑎 𝑥→𝑎
lim [𝑓 𝑥 − 𝑔(𝑥)]
𝑥→𝑎
is called an indeterminate form of type ∞ − ∞.

To find out, we try to convert the difference into a quotient, by

using:
▪ a common denominator,
▪ or rationalization,
▪ or factoring out a common factor
𝟎
so that we have an indeterminate form of type or ∞Τ∞ so
𝟎
that we can use l’Hopital’s Rule.
Example 5
1 1
Compute lim+
ln 𝑥
− 𝑥−1
.
𝑥→1
Solution
Clearly the limit has the form ∞ − ∞, since
1 1
lim+ =∞ and lim+ =∞
𝑥→1 ln 𝑥 𝑥→1 𝑥−1
Here we can start with a common denominator:

Both numerator and denominator have a limit of 0, so l’Hopital’s

Rule applies, giving
 𝟎
Again we have an indeterminate limit of type , so we apply
𝟎
l’Hopital’s Rule a second time:
Example 6
Calculate lim (𝑥 − ln 𝑥) .
𝑥→∞
Solution
The limit has the form ∞ − ∞, and we will change the form to
a product by factoring out 𝑥
(Note: we factor out the function that goes to ∞ most rapidly )
Hence,
ln 𝑥
lim (𝑥 − ln 𝑥) = lim 𝑥 1 −
𝑥→∞ 𝑥→∞ 𝑥
ln 𝑥
Now compute the limit of 𝑥
using l’Hopital’s Rule:
ln 𝑥 𝐻 1Τ𝑥 1
lim ֜ = lim = lim = 0
𝑥→∞ 𝑥 𝑥→∞ 1 𝑥→∞ 𝑥
Consequently,
ln 𝑥
lim (𝑥 − ln 𝑥) = lim 𝑥 1 −
𝑥→∞ 𝑥→∞ 𝑥
ln 𝑥
= lim 𝑥 . lim 1 −
𝑥→∞ 𝑥→∞ 𝑥
= ∞ 1 − 0 = ∞.
Case 4: Indeterminate Forms 𝟎𝟎 , ∞𝟎 & 𝟏∞
Several indeterminate forms arise from the limit
𝑔(𝑥)
lim 𝑓(𝑥)
𝑥→𝑎
▪ lim 𝑓 𝑥 = 0 and lim 𝑔 𝑥 = 0 type 𝟎𝟎
𝑥→𝑎 𝑥→𝑎
▪ lim 𝑓 𝑥 = ∞ and lim 𝑔 𝑥 = 0 type ∞𝟎
𝑥→𝑎 𝑥→𝑎
▪ lim 𝑓 𝑥 = 1 and lim 𝑔 𝑥 = ∞ type 𝟏∞
𝑥→𝑎 𝑥→𝑎

Each of these three cases can be treated by taking the natural

logarithm:
Let y = 𝑓(𝑥) 𝑔(𝑥)
then, lim ln 𝑦 = lim 𝑔 𝑥 ln 𝑓(𝑥)
𝑥→𝑎 𝑥→𝑎
which in turn leads us to the indeterminate product
𝑔 𝑥 ln 𝑓(𝑥) , which is of type 𝟎. ∞.
Example 7
cot 𝑥
Calculate lim+ 1 + sin 4𝑥 .
𝑥→0
Solution
First notice that
lim + 1 + sin 4𝑥 = 1 and lim+ cot 𝑥 = ∞
𝑥→0 𝑥→0

so the given limit is indeterminate (type 𝟏∞ ).

Let 𝑦 = 1 + sin 4𝑥 cot 𝑥

then, ln 𝑦 = ln 1 + sin 4𝑥 cot 𝑥 = cot 𝑥 ln(1 + sin 4𝑥 )

Now we rewrite cot 𝑥 ln(1 + sin 4𝑥 ) as
ln(1+sin 4𝑥 )
cot 𝑥 ln(1 + sin 4𝑥 ) = (why?)
tan 𝑥
so l’Hopital’s Rule gives

So far we have computed the limit of ln 𝑦, but what we want is

the limit of 𝑦. To find this we use the fact that 𝑦 = 𝑒 ln 𝑦 :
Thank you for listening.