- L'Hopital's rule provides a way to evaluate limits that result in indeterminate forms like 0/0 or ∞/∞ by taking the derivative of the top and bottom functions. - It can be applied repeatedly as long as the limit remains indeterminate. You stop differentiating once it is no longer indeterminate. - Some indeterminate forms like 1∞ can be rewritten as ratios using properties like ln(u^n) = n*ln(u) to apply L'Hopital's rule. - Examples are provided to demonstrate evaluating limits of various indeterminate forms using L'Hopital's rule, including its repeated application and stopping early if the form

1. L’HOSPITAL’S RULE
Presented By:
Balaj Khan
Syed kumail

2. Guillaume De l'Hôpital
1661 - 1704
Indeterminate forms
and L’Hospital’s Rule

3. Zero divided by zero can not be evaluated. The limit may
or may not exist, and is called an indeterminate form.
2
2
4
lim
2x
x
x→
−
−
Consider: or
If we try to evaluate by direct substitution, we get:
0
0
In the case of the first limit, we can evaluate it by
factoring and canceling:
2
2
4
lim
2x
x
x→
−
−
( ) ( )
2
2 2
lim
2x
x x
x→
+ −
=
−
( )2
lim 2
x
x
→
= + 4=
1
ln
lim1 −→ x
x
x
This method does not work in the case of the second limit.
Indeterminate forms

4. Suppose f and g are differentiable and g’(x) ≠ 0 near a (except possible at
a). Suppose that
L’Hospital’s Rule
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limlim
limlim
limlim
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0
0
−∞→∞→→→
→
∞∞
′
′
=
∞∞
±∞=±∞=
==
+−
→→
→→
→→
xxaxax
xg
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xg
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xgxf
axax
axax

5. L’Hospital’s Rule: Examples
2
2
4
lim
2x
x
x→
−
=
−
( )
( )
lim
x a
f x
g x→
( )
( )
2
2
4
lim
2
x
d
x
dx
d
x
dx
→
−
=
−
2
2
lim
1x
x
→
= 4=
1
1
1
1
)1(
)(ln
1
ln
limlimlimlim 1111
===
−
=
− →→→→ x
x
x
dx
d
x
dx
d
x
x
xxxx
∞===
∞→∞→∞→ 1)(
)(
limlimlim
x
x
x
x
x
x
e
x
dx
d
e
dx
d
x
e

6. Example:
20
1 cos
lim
x
x
x x→
−
+ 0
sin
lim
1 2x
x
x→
=
+
0=
If it’s no longer
indeterminate, then
STOP differentiating!
If we try to continue with L’Hôpital’s rule:
0
sin
lim
1 2x
x
x→
=
+ 0
cos
lim
2x
x
→
=
1
2
=
which is wrong!
→

7. On the other hand, you can apply L’Hôpital’s rule as
many times as necessary as long as the fraction is still
indeterminate:
20
1 1
2lim
x
x
x
x→
+ − −
( )
1
2
0
1 1
1
2 2lim
2x
x
x
−
→
+ −
=
0
0
0
0
0
0
not
( )
1
2
20
1
1 1
2lim
x
x x
x→
+ − −
( )
3
2
0
1
1
4lim
2x
x
−
→
− +
=
1
4
2
−
= 1
8
= −

8. 1
lim sin
x
x
x→∞
 
 
 
This approaches
0
0
1
sin
lim
1x
x
x
→∞
This approaches 0∞⋅
We already know that 0
sin
lim 1
x
x
x→
 
= 
 
but if we want to use L’Hôpital’s rule:
2
2
1 1
cos
lim
1x
x x
x
→∞
   
⋅ −   
   =
−
1
sin
lim
1x
x
x
→∞
1
limcos
x x→∞
 
=  
 
( )cos 0= 1=
Indeterminate Products
Rewrite as a ratio!

9. 1
1 1
lim
ln 1x x x→
 
− 
− 
If we find a common denominator and subtract, we get:
( )1
1 ln
lim
1 lnx
x x
x x→
 − −
  − 
Now it is in the form
0
0
∞ − ∞This is indeterminate form
1
1
1
lim
1
ln
x
x
x
x
x
→
 
− 
 −
 + 
 
L’Hôpital’s rule applied once.
0
0
Fractions cleared. Still1
1
lim
1 lnx
x
x x x→
− 
 ÷
− + 
Indeterminate Differences
Rewrite as a ratio!
1
1
lim
1 1 lnx x→
 
 ÷
+ + 
L’Hôpital again.
1
2
Answer:

10. Indeterminate Forms: 1∞ 0
0 0
∞
Evaluating these forms requires a mathematical trick to
change the expression into a ratio.
ln lnn
u n u=
ln
1
u
n
=
We can then write the
expression as a ratio,
which allows us to use
L’Hôpital’s rule.
( )lim
x a
f x
→
( )( )ln lim
x a
f x
e →
=
( )( )lim ln
x a
f x
e →
=
We can take the log of the function as long
as we exponentiate at the same time.
Then move the
limit notation
outside of the log.
→
Indeterminate Powers

11. Indeterminate Forms: 1∞ 0
0 0
∞
1/
lim x
x
x
→∞
( )1/
lim ln x
x
x
e →∞
( )
1
lim ln
x
x
x
e →∞
( )ln
lim
x
x
x
e →∞
1
lim
1x
x
e →∞
0
e
1
0
∞
∞
∞
L’Hôpital
applied
Example: