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Guillaume De l'Hpital

1661 - 1704

LHpitals Rule

Actually, LHpitals Rule wasdeveloped by his teacher Johann Bernoulli. De lHpital

paid Bernoulli for privatelessons, and then published thefirst Calculus book based on

those lessons.

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Johann Bernoulli

1667 - 1748

LHpitals Rule

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Zero divided by zero can not be evaluated, and is anexample of an indeterminate form .

2

2

4lim

2 x

x

xConsider:

If we try to evaluate this by direct substitution, we get:00

In this case, we can evaluate this limit by factoring andcanceling:

2

2

4lim

2 x x x

2

2 2lim

2 x x x

x

2

lim 2 x

x 4

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If we zoom in far enough,the curves will appear as

straight lines.

2

2

4lim

2 x

x

xThe limit is the ratio of the numerator

to the denominator

as

x approaches 2.

-5

-4

-3

-2

-10

1

2

3

4

-3 -2 -1 1 2 3x

2 4 x

2 x -0.05

0

0.05

1.95 2 2.05x

lim x a

f x

g x

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2

2

4lim

2 x

x

x

lim x a

f x

g x

-0.05

0

0.05

1.95 2 2.05x

f x

g x

f x

g x

As 2 x

becomes:

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2

2

4lim

2 x

x

x

lim x a

f x

g x

-0.05

0

0.05

1.95 2 2.05x

As 2 x

f x

g x

becomes:

df

dg

df dg

dx

dx

df

xdgd

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2

2 4lim 2 x x x lim x a f x

g x

2

2

4lim

2 x

d x

dxd x

dx

22lim1 x x 4

LHpitals Rule:

If is indeterminate, then:

lim x a

f x

g x

lim lim x a x a

f x f x

g x g x

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We can confirm LHpitals rule by working backwards, and

using the definition of derivative:

f a

g a

lim

lim

x a

x a

f x f a

x ag x g a

x a

lim x a

f x f a

x ag x g a

x a

lim x a

f x f a

g x g a

0lim

0 x a f x

g x

lim x a

f x

g x

Let f and g be defined on an interval where f(x) 0 and

g(x) 0 as x a, but tends to a finite limit L.

xg

x f

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Example:

20

1 coslim x

x x x 0

sinlim

1 2 x x x

0

If its no longer indeterminate, thenSTOP !

If we try to continue with LHpitals rule:

0

sinlim

1 2 x

x

x0

coslim

2 x

x 1

2

which is wrong,

wrong, wrong!

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On the other hand, you can apply LHpitals rule asmany times as necessary as long as the fraction is still

indeterminate:

20

1 12lim

x

x x

x

12

0

1 11

2 2lim 2 x

x

x

0

0

0

0

0

0not

12

20

11 1

2lim x

x x

x

32

0

11

4lim 2 x

x 14

2

1

8

(Rewritten inexponentialform.)

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When applying LHpitals

Rule, one has to differentiate the

numerator and the denominator separately .

A common mistake is to differentiate the whole expression f(x

) / g(x)

This leads to tedious, unnecessary, and, most importantly,

wrong computations.

Correct calculation:

2

1

2cos2lim

2sinlim

2sinlim

000

x

xdxd

xdxd

x

x x x x

Incorrect calculation:

????2sin2cos2lim2sinlim 200 x x x x

x x

x x

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LHpitals rule can be used to evaluate other indeterminate

00

forms besides .

The following are also considered indeterminate:

0 10

00

The first one, , can be evaluated just like .0

0The others must be changed to fractions first.

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What makes an expression indeterminate?

lim 1000 x x

Consider:

We can hold one part of the expression constant:

1000lim 0 x x

There are conflicting trends here. The actual limitwill depend on the rates at which the numerator anddenominator approach infinity, so we say that anexpression in this form is indeterminate .

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Lets look at another one:

0

0lim1000 1 x

x

Consider:

We can hold one part of the expression constant:

0.1

lim x x

Once again, we have conflicting trends, so this formis indeterminate .

0.1

lim 0 x x

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Here is an expression that looks like it might be

indeterminate :

0

lim .1 0 x

x

Consider:

We can hold one part of the expression constant:

lim .1 0 x

x

The limit is zero any way you look at it, so theexpression is not

indeterminate .

1000

0lim 0 x x

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Finally, here is an expression that looks like it

should NOT be indeterminate :

We can hold one part of the expression constant:

Consider: 1

461000411000

100010001000

-101.759.0;102.471.1

:

01

lim;11;1

lim

ex

x x

x x

Once again, we have conflicting trends, so this form is

indeterminate .

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Here is the standard list of indeterminate forms:

0

1 00 0

00

There are other indeterminate forms using complexnumbers, but those are beyond the scope of this class.

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Example

x 1

1 1

x 1 1

Let's return to our former example: lim1

1. Check to see if L'Hopital's rule applies:lim( ) 0 and lim( 1) 0

L'Hopital's rule does apply.

( )lim lim lim

1 ( 1)

x

x

x x

x x

x

e e x

Stepe e x

d e ee e dx

d x xdx

1

1

x

x

ee

a

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Example:

1lim

1lim

1lim 2

2

2

2

2 x x

x

x

x

x x x x

1lim

2 x

x x

;1

lim1

lim 22

2

2

x x

x x

x x

Determine the limit:

Rewrite:

Use the Power Law: The limit of a functions positive integer power is the power of the functions limit:

OR Ratio of the leading powers:

1

1lim

1lim

2

2

2

2

xdxd

xdxd

x x

x x

111

1lim

1lim 2

2

2

2

x x

x x

x x

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1lim sin x

x x

This approaches00

1sin

lim

1 x

x

x

This approaches 0

We already know that0

sinlim 1 x

x

x

but if we want to use LHpitals rule:

2

2

1 1cos

lim1 x

x x

x

1sinlim

1 x x

x

1lim cos x x

cos 0 1

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1

1 1lim

ln 1 x x x

If we find a common denominator and subtract, we get:

11 lnlim1 ln x

x x x x

Now it is in the form00

This is indeterminate form

1

11lim

1ln

x

x x

x

x

LHpitals rule applied once.

0

0

Fractions cleared. Still1

1lim

1 ln x x

x x x

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1

1 1lim

ln 1 x x x

11 lnlim1 ln x

x x x x

1

11lim

1ln

x

x x

x x

1

1lim 1 1 ln x x

LHpital

again.

1

2

1

1lim

1 ln x

x

x x x

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Cautionary Example

1

1 1

1

lnExample: Evaluate lim

Step 1. Check to see if L'Hopital/s rule applies:

lim ln ln1 0 but lim 1 0

L'Hopital's Rule does not apply.

Use the quotient property for limits instead:limln

lim

x

x x

x

x

x x

x x

x x

1

1

ln 00

lim 1

Using L'Hopital's Rule would have given us

an incorrect result.

x

x

x

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Indeterminate Forms: 1 00 0

Evaluating these forms requires a mathematical trick tochange the expression into a fraction.

ln lnnu n u

When we take the log of an exponential function,the exponent can be moved out front.

ln1u

n

We can then write theexpression as a fraction,which allows us to useLHpitals

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 thelimit notationoutside of the log.

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Indeterminate Forms: 1 00 0

1/lim x x

x

1/lim ln x x xe

1lim ln x

x xe

lnlim

x

x

x

e1

lim 1 x x

e0

e1

0

LHpitalapplied

Example: