This slide introduces you with Indeterminate forms and how to solve equations and limits using L'Hospital's Rule

1. Sikkim Institute of Science &
Technology
Mathematics Presentation
By – Bijay Sharma
1st year - Computer Science & Engineering Department

2. Indeterminate forms
and L’Hospital’s Rule

3. Contents
✘ What do we mean by indeterminate
forms?
✘ Forms of Indeterminates
✘ Why L’Hosipital’s Rule?
✘ Understanding L’Hospital’s Rule
✘ Example Problem using L’Hospital’s
Rule
✘ 4 Problems with L’Hospital’s Rule
3

4. What do we mean by Indeterminate form?
Something is said to be of indeterminate when
it has no fixed numeric value.
Example:
0
0
,
∞
∞
, ∞ − ∞, 0°, 1∞
, ∞0
, 0 × ∞

5. Big concept
Infinity (∞) is not a number,
rather, it exists only as an
abstract concept.
5

6. 6
8
2
= 4
∴ 𝐼𝑡 𝑖𝑠 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
0
0
= 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
∞
∞
= 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
NB :
∞
∞
= 1
∞ − ∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
00
= 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
1∞
= 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
∞0
= 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
0 × ∞ = 𝑛𝑜 𝑓𝑖𝑥𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
∴ 𝐼𝑡 𝑖𝑠 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒
Indeterminate
forms

7. Why do we need L’Hopital’s Rule ?
Let’s say we want to evaluate the given limit
lim
𝑥→0
sin 𝑥
𝑥
=
0
0
This limit is indeterminate
7

8. L’Hopital
Guillaume François
Antoine, Marquis de
l'Hôpital
8

9. L’Hospital’s Rule
L'Hospital's rule uses derivatives to help evaluate limits involving
indeterminate forms. Application of the rule often converts an
indeterminate form to an expression that can be evaluated by
substitution, allowing easier evaluation of the limit.

10. L’Hospital’s Rule
Suppose that we have one of the following cases:
lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
0
0
𝑜𝑟 lim
𝑥→𝑎
𝑓(𝑥)
𝑔(𝑥)
=
±∞
±∞
𝑤ℎ𝑒𝑟𝑒 𝑎 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦 𝑜𝑟 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦.
𝐼𝑛 𝑡ℎ𝑒𝑠𝑒 𝑐𝑎𝑠𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒,
lim
𝑥→𝑎
𝑓(𝑥)
𝑔 𝑥
= lim
𝑥→𝑎
𝑓′(𝑥)
𝑔′(𝑥)
So, L’Hospital’s Rule tells us that if we have an indeterminate form
or all we need to do is differentiate the numerator and differentiate
the denominator and then take the limit.
10

11. Example
11
Let’s say we want to find the solution for 𝐥𝐢𝐦
𝒙→𝟎
𝐬𝐢𝐧 𝒙
𝒙
lim
𝑥→0
sin 𝑥
𝑥
=
0
0
(but, this limit is indeterminate)
∴ 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑙𝑦 𝐿′
𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′
𝑠 𝑟𝑢𝑙𝑒
lim
𝑥→0
sin 𝑥
𝑥
= lim
𝑥→0
cos 𝑥
𝑥
=
1
1
= 1

12. 12
Exercise

13. Problem 1 : Evaluate the given limit : lim
𝑡→1
5𝑡4−4𝑡2−1
10−𝑡−9𝑡³
lim
𝑡→1
5𝑡4
− 4𝑡2
− 1
10 − 𝑡 − 9𝑡3
=
5(1)4
− 4(1)2
−1
10 − 1 − 9(1)³
=
0
0
We get an indeterminate form.
Hence we will apply L’Hospital’s rule to evaluate the limit
⇒ lim
𝑡→1
5𝑡4
− 4𝑡2
− 1
10 − 𝑡 − 9𝑡³
= lim
𝑡→1
20𝑡3
− 8𝑡
−1 − 27𝑡²
= −
3
7
13

14. Problem 2 : Evaluate the given limit : lim
𝑥→∞
𝑒 𝑥
𝑥³
lim
𝑥→∞
𝑒 𝑥
𝑥²
=
∞
∞
(indeterminate)
∴ 𝑤𝑒 𝑎𝑝𝑝𝑙𝑦 𝐿′
𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′
𝑠 𝑟𝑢𝑙𝑒
lim
𝑥→∞
𝑒 𝑥
𝑥²
= lim
𝑥→∞
𝑒 𝑥
2𝑥
(The new limit also turns out to be indeterminate)
We know how to deal with these kinds of limits, We will just apply L’Hospital’s rule again
lim
𝑥→∞
𝑒 𝑥
𝑥²
= lim
𝑥→∞
𝑒 𝑥
2𝑥
=
𝑒 𝑥
2
= ∞
NB – We can apply L’Hospital’s rule more than once to get the
results.
14

15. Problem 3 : Evaluate the given limit : lim
𝑥→0
𝑥+sin 𝑥
𝑥+cos 𝑥
lim
𝑥→0
𝑥+sin 𝑥
𝑥+cos 𝑥
=
0
1
= 0
Using the L’Hospital’s rule here would have given us the
wrong answer.
The rule only works on indeterminate forms.
15

16. Big concept
lim
𝑥→0
𝑓 𝑥 . 𝑔(𝑥) = (0)(∞)
lim
𝑥→0
𝑓 𝑥 . 𝑔(𝑥) = lim
𝑥→0
𝑓(𝑥)
1 𝑔(𝑥)
= lim
𝑥→0
𝑔(𝑥)
1 𝑓(𝑥)
Now we can apply L’Hospital’s Rule
16

17. Problem 4 : Evaluate the given limit : lim
𝑥→0+
𝑥. ln 𝑥
lim
𝑥→0+
𝑥. ln 𝑥 = (0)(∞) (indeterminate)
= lim
𝑥→0+
ln 𝑥
𝑥−1
Now lets take the derivative
= lim
𝑥→0+
1
𝑥
−𝑥−2
= lim
𝑥→0+
1
𝑥
.
−𝑥²
1
= lim
𝑥→0+
− 𝑥 = 0
17

18. Problem 5 : Evaluate the given limit : lim
𝑥→∞
4𝑥2−5𝑥
1−3𝑥²
lim
𝑥→∞
4𝑥2−5𝑥
1−3𝑥²
=
∞
−∞
(indeterminate)
We will apply L’Hospital’s rule.
= lim
𝑥→∞
8𝑥 − 5
−6𝑥
=
∞
−∞
NB – We can apply L’Hospital’s rule more than once.
∴ 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑎𝑝𝑝𝑙𝑦 𝐿′
𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙′
𝑠 𝑟𝑢𝑙𝑒 𝑎𝑔𝑎𝑖𝑛
= lim
𝑥→∞
8
−6
=
𝟒
−𝟑
Bonus – Let’s see another way to solve the same problem in the
next slide.18

19. Bonus : Evaluate the given limit : lim
𝑥→∞
4𝑥2−5𝑥
1−3𝑥²
lim
𝑥→∞
4𝑥2−5𝑥
1−3𝑥²
=
∞
−∞
(indeterminate)
= lim
𝑥→∞
𝑥²(4 − 5
𝑥)
𝑥²(1
𝑥2 − 3)
= lim
𝑥→∞
4 − 5
𝑥
1
𝑥²
− 3
=
𝟒
−𝟑
19

20. 20
The End
Thank You ! 😊