L’Hôpital’s Rule

IB HL Mathematics Internal Assessment

Word Count : 2207

1

Table of Contents

Introduction 2

Background Knowledge on Limits 3

Proof of L’Hôpital’s Rule 6

Applying L’Hôpital’s Rule 7

History of L’Hôpital’s Rule 9

Conclusion and Reflection 10

Works Cited 11

2

Introduction

I have loved numbers and math for as long as I can remember. When I started

school, math was always my favourite subject. I would even ask my mother to give me

math problems at home, simply because I loved doing them. I wanted to pick an IA topic

that interested me on a personal level, and that personal level is definitely math.

Everyday in IB HL Math class, I love learning new equations, rules and tricks to do

something new. When we were told to pick our IA topics, we needed to link math with

something you like. Well I like math. How do you link math with more math? I then

decided that I wanted to learn a new rule or law and learn its proof.

I first learned about limits when my class was taught how to derive using first

principles. I thought limits were quite fascinating because even though the number

never reaches zero, we label it as zero anyway. When I approach a wall, even if I am

squished up against it… no one will ever say “Oh, she is close enough to the wall, so

she is now a wall”. But numbers work differently than actual 3D objects. As I am

scrolling through a list of possible IA topics, one caught my eye. Maybe it intrigued me

because it looked like the ‘ô’ was wearing a hat; or maybe it was because it was french

and I was excited about nearly finishing my many french assignments; or maybe it was

just destiny that I decided to click on l’Hôpital’s rule. L’Hôpital’s rule shares my new

found love for limits and deriving, as well as my interest in things that look like they are

impossible, but are not. Once I chose my topic, I watched a lot of neat videos on how

l’Hôpital’s rule worked and it was like I was transported into the world of calculus, where

people loved math just as much as I do!

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Background Knowledge on Limits

A limit is the value that a function approaches as the input, approaches somex

value, . When approaches from the right, it is which is called a right handa x a (x) limx→a+

f

limit. Right hand limits are when approaches through values that are greater than x a a

. When approaches from the left it is which is called a left hand limit. Leftx a (x)limx→a−

f

hand limits are when approaches through values that are less than . For a limit tox a a

exist, both (the right hand limit) and (the left hand limit) must exist and(x) limx→a+

f (x)limx→a−

f

be equal to each other.

In example 1: on the left, is a simple(x )limx→6

+ 2 = 8

case where can be evaluated directly, but when we(6)f

calculate limits we look at being close enough to tox 6

label it as since it is so close.6

Example 2 on the right is of the function

. The limit, as it approaches does not(x)f = 1x−3 3

exist. The red arrows are and the blue(x)limx→3−

f

arrows are . The arrows along the function(x)limx→3+

f

are not pointing in the same direction. The red

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arrow is going towards and the blue arrow is going towards . Since− ∞ + ∞

then does not exist. I included these examples to try and(x) = (x)limx→3−

f / limx→3+

f (x)limx→3

f

give some background knowledge of what limits are because limits are necessary for

learning l’Hôpital’s rule.

In example 3, there is alimx→−1 x+1

x +3x+22

discontinuity at . This means, that there is a−x = 1

spot on the function that does not have a value.

There is a void in the line. When plugging in as− 1

, you get that the limit is which does not exist.x 00

But we know that there is a limit and that it can be

solved by using l’Hôpital’s rule. For now we can simply look at the graph and infer that

the limit as approaches equals .x − 1 1

This is when l’Hôpital’s rule comes into play. L’Hôpital’s rule is a rule that is

applied when attempting to evaluate a limit at the answer is or , these area 00

±∞±∞

indeterminate forms that can not be defined as any value. The rule is, that when you

have a limit that equals or , you differentiate the numerator and the denominator.00

±∞±∞

After differentiating, you evaluate the limit as normal, until you get a real value as an

answer. Sometimes you have to differentiate multiple times to get an answer that isn’t 00

or . Since division by zero is impossible, I wondered how it is even possible to get±∞±∞

another answer? How can you have as one answer and magically have a real answer00

once applying l’Hôpital’s rule? After deriving both numerator and denominator in

5

you get which can be evaluated and once you plug in in as limx→−1 x+1

x +3x+22 limx→−1 1

2x+3 − 1 x

you get as the limit, which is what it is on the graph too.1

The same thought process happens when thinking of , infinity is not a±∞±∞

constant. There is not a set number that infinity ends on, that you can use for this

fraction. Putting common knowledge into place, I wondered if it was possible that ∞∞ = 1

and . But as famous author John Green once wrote, “Some infinities are bigger∞−∞ = − 1

than other infinities”. So is it even possible to compare one infinity to the other? And is it

wrong to assume that they would obtain the same value? For my sake, I will only be

proving the case where the limit is equal to , because the case is too complicated.00

±∞±∞

6

Proof of L’Hôpital’s Rule

The general form of a limit is . Let’s keep in mind that the derivative of alimx→a

f(x)g(x)

function at the point where , showed by is given by . Now(x)f x = a (x)f ′ (a)f ′ = limx→a x−a

f(x)−f(a)

since this is the case of the rule, when we plug in into the limit as we can say that00 a x

and . In this proof, , and are both continuous and(a)f = 0 (a)g = 0 (a) =g′ / 0 (x)f ′ (x)g′

. When a function is continuous it means that there is not a break in the line, so if=x / a

there was a part missing it would be a discontinuous function.

limx→a

f(x)g(x) = lim

x→af(x)−f(a)g(x)−g(a) (since )(a) (a)f = g = 0

= limx→a

x−af (x)−f (a)

x−ag(x)−g(a) (multiplied by because )1

x−a

1x−a = 1 =x / a

=limx→a x−a

f (x)−f (a)

limx→a x−a

g(x)−g(a) (limit laws)

= f (a)′g (a)′ (because of the structure of a derivative shown above)

=(x)lim

x→af ′

(x)limx→a

g′ (because and are continuous)(x)f ′ (x)g′

= limx→a

f (x)′g (x)′ Which proves l’Hôpital’s rule as .lim

x→af(x)g(x) = lim

x→af (x)′g (x)′

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Applying L’Hôpital’s Rule

Let’s solve a few other limits by using l’Hôpital’s rule.

Example 4: limx→0 x2

e −1−xx

=02

e −1−00(first filling in as )x 0

= 00 (since the limit is we must apply l’Hôpital’s rule)0

0

= limx→0 2x

e −1x(now using the derivatives of the numerator and denominator)

= 2(0)e −10

(filling in as again)x 0

= 00 (since we got we need to apply l’Hôpital’s rule again)0

0

= limx→0 2

ex (now deriving the numerator and denominator again)

= 2e0

(filling in as )x 0

= 21

Therefore the limit as approaches for the equation equals .x 0 x2e −1−xx

21

Example 5: limx→2

x −x −x−23 2

x −3x +3x−23 2

= 2 −2 −2−23 2

2 −3(2) +3(2)−23 2 (filling in because is so close to it that we label it as )2 x 2

= 8−4−2−28−4−2−2

= 00 (indeterminate form therefore apply l’Hôpital’s rule)

= limx→2 3x −6x+32

3x −2x−12(deriving numerator and denominator)

= 3(2) −6(2)+323(2) −2(2)−12

(fill in again)2

= 12−4−112−12+3

= 37

So the limit as approaches for the equation equals .x 2 x −x −x−23 2

x −3x +3x−23 2 37

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Example 6, let’s try using l’Hôpital’s rule as approaches : x + ∞ limx→ +∞ 2x +4x+92

7x −6x+22

= 2∞ +4∞+927∞ −6∞+22

(filling in as because it is approaching )∞ x ∞

= ∞∞ (indeterminate form, therefore apply l’Hôpital’s rule)

= limx→+∞ 4x+4

14x−6 (deriving numerator and denominator)

= 4∞+414∞−6 (filling in as again)∞ x

= ∞∞ (another indeterminate form)

= limx→+∞ 4

14 (deriving numerator and denominator again)

= 27

Therefore the limit as approaches for the equation equals .x + ∞ 2x +4x+927x −6x+22

27

I included these examples to show that sometimes it is needed to use l’Hôpital’s

rule more than once to get the correct limit. I also wanted to show at least one example

where approaches because it is a large part in l’Hôpital’s rule, even if I did notx ∞

prove that element of the rule in my IA. Mostly I wanted to show how l’Hôpital’s is

applied and how it works to evaluate a limit that has an indeterminate form.

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History of L’Hôpital’s Rule

Although the name clearly states that this is the rule of L’Hôpital,

Guillaume-François-Antoine Marquis de l'Hôpital was not the one who discovered the

rule. He had made an agreement with a man named Johann Bernoulli that L’Hôpital

could use Bernoulli’s discoveries as he pleased, and so he did. Since Bernoulli was the

one who first discovered the rule and L’Hôpital was first to publish it, many claim that he

stole the rule that was not rightfully his. Many other people who hear this story claim

L’Hôpital bought his way into the world of being a mathematician. But that is simply not

the case.

L’Hôpital published the rule in 1696 in the first textbook on differential calculus,

“L’Analyse des infiniment petits pour l’intelligence des lignes courbes”. In the textbook,

l’Hôpital thanks Bernoulli for his assistance and his discoveries. Also L’Hôpital never

called the rule by his own name. It was not until many years later that the rule was

called ‘L’Hôpital’s rule’. So while these people fight over how undeserving L’Hôpital is of

the rule, there is no evidence that Bernoulli was even upset at all over what happened.

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Conclusion and Reflection

Having completed this internal assessment, I feel that I have achieved my main

goal, which was learning a new rule and its proof. When I first started my research, I

was easily confused by the math used in certain examples of l’Hôpital’s rule. As I

continued to work through the assessment, I became more confident in my knowledge

of the rule. I feel that I made the right decision by choosing to only include the proof for

the case in my IA. I think it would have been too complicated to include the proof. I00

±∞±∞

am pleased with how the examples turned out in this assessment. By drawing the

examples I was able to show the readers exactly what I wanted to show them. I am glad

that I chose l'Hospital's rule as a topic because being the person I am, I do not think

another topic could have interested me as much as this one. I had no prior knowledge

on l'Hospital's rule before I wrote this and I genuinely enjoyed writing this paper. As I

reflect on the experience of this internal assessment, I have realized the importance of

independent research and know that what I have learned will be beneficial to me in

future assignments.

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Works Cited

Ayres, Frank, and Elliott Mendelson. Calculus . 4th ed. Print. Schaum's Outline.

"Calculus I - L'Hospital's Rule and Indeterminate Forms." Calculus I - L'Hospital's Rule

and Indeterminate Forms . Web. 05 June 2016.

"History." L'Hospital's Rule . Web. 08 June 2016.

"Introduction to L'Hôpital's Rule." Khan Academy . Web. 05 June 2016.

"L'Hospital's Rule." -- from Wolfram MathWorld . Web. 05 June 2016.

"L'Hopital's Rule." Encyclopedia Britannica Online . Encyclopedia Britannica, Web. 08

June 2016.

Quinn, Catherine, C. J. Sangwin, R. C. Haese, and Michael Haese. Mathematics for the

International Student: Mathematics HL (option): Calculus, HL Topic 9, FM Topic

5, for Use with IB Diploma Programme . Print.

"The Rule of L'Hospital." L'Hopital's Rule . Web. 08 June 2016.