LESSON 64 L’HOPITAL’S RULE HL MATH - CALCULUS Objective: To use L’Hopital’s Rule to solve problems involving limits

LESSON 64 L’HOPITAL’S RULE

HL MATH - CALCULUS

Objective: To use L’Hopital’s Rule to solve problems involving limits.

EXAMPLES TO START WITH

€

limx→ 2

x2 −4x−2

limx→ 3

x+1 −2x−3

limx→ 3

2x−6x2 −9

limx→ ∞

5x−27x+3

limx→ ∞

3x2 +5x−72x2 −3x+1

limx→ ∞

4x2 +1x+3

Given our previous work with limits, evaluate the following limits:

CHALLENGE EXAMPLE

So how would we evaluate

€

limx→ −2

x+2(ln x+3)

FORMS OF TYPE 0/0

We looked at several limits in chapter 2 that had the form 0/0. Some examples were:

1sin

lim0

x

xx

21

1lim

2

1

x

xx

0cos1

lim0

x

xx

FORMS OF TYPE 0/0

We looked at several limits in chapter 2 that had the form 0/0. Some examples were:

The first limit was solved algebraically and the next two were solved using geometric methods. There are other types that cannot be solved by either of these methods. We need to develop a more general method to solve such limits.

1sin

lim0

x

xx

21

1lim

2

1

x

xx

0cos1

lim0

x

xx

FORMS OF TYPE 0/0

Lets look at a general example. 00

)(

)(lim xg

xfax

FORMS OF TYPE 0/0

Lets look at a general example.

We will assume that f / and g / are continuous at x = a and g / is not zero. Since f and g can be closely approximated by their local linear approximation near a, we can say

))(()(

))(()(lim

)(

)(lim

/

/

axagag

axafaf

xg

xfaxax

00

)(

)(lim xg

xfax

FORMS OF TYPE 0/0


We know that:

So we can now say

))((

))((

))(()(

))(()(lim

)(

)(lim

/

/

/

/

axag

axaf

axagag

axafaf

xg

xfaxax

00

)(

)(lim xg

xfax

0)(lim)(

xfafax

0)(lim)(

xgagax

FORMS OF TYPE 0/0


This will now become

This is know as L’Hopital’s Rule.

)(

)(

)(

)(

)(

)(lim

/

/

/

/

xg

xf

ag

af

xg

xfax

00

)(

)(lim xg

xfax

THEOREM 4.4.1

L’Hopital’s Rule: Suppose that f and g are differentiable functions on an open interval containing x = a, except possibly at x = a, and that

and

If exists or if this limit is then )](/)([lim // xgxfax

0)(lim

xfax

.0)(lim

xgax

)(

)(lim

)(

)(lim

/

/

xg

xf

xg

xfaxax

EXAMPLE 1

Find the following limit. 2

4lim

2

2

x

xx

EXAMPLE 1

Find the following limit.

Now, we can use L’Hopital’s Rule to solve this limit.

2

4lim

2

2

x

xx

4)2(21

2

2

4lim

2

2

x

x

xx

EXAMPLE 2

Find the following limit. x

xx

2sinlim

0

EXAMPLE 2


First, we confirm that this of the form 0/0. Next, we use L’Hopital’s Rule to find the limit.

21

)1(2

1

2cos22sinlim

0

x

x

xx

x

xx

2sinlim

0

EXAMPLE 2

Find the following limit. x

xx cos

sin1lim

2/

EXAMPLE 2



x

xx cos

sin1lim

2/

01

0

sin

cos

cos

sin1lim

2/

x

x

x

xx

EXAMPLE 2


1lim

x

ex

x

EXAMPLE 2



This is an asymptote. Do sign analysis to find the answer. ___+___|___+___

0

0

1

3

1lim

230 x

e

x

e xx

x

30

1lim

x

ex

x

EXAMPLE 2


tanlim

x

xx

EXAMPLE 2


First, we confirm that this of the form 0/0. Next we use L’Hopital’s Rule to find the answer.

Again, sign analysis. ___-___|___+___ 0

20

tanlim

x

xx

0

1

2

sectanlim

2

20 x

x

x

xx

EXAMPLE 2


cos1lim

x

xx

EXAMPLE 2



We apply L’Hopital’s Rule again.

0

0

2

sincos1lim

20

x

x

x

xx

20

cos1lim

x

xx

EXAMPLE 2



We apply L’Hopital’s Rule again.

0

0

2

sincos1lim

20

x

x

x

xx

20

cos1lim

x

xx

2

1

2

cos

2

sincos1lim

20

x

x

x

x

xx

EXAMPLE 2

Find the following limit. )/1sin(lim

3/4

x

xx

EXAMPLE 2



)/1sin(lim

3/4

x

xx

01

0

)/1cos(

)3/4(

)/1cos()/1(

)3/4(

)/1sin(lim

3/1

2

3/73/4

x

x

xx

x

x

xx

EXAMPLE 2



04

)/1cos(3

4

)/1cos()/1(

)3/4(

)/1sin(lim

3/12

3/73/4

xxxx

x

x

xx

)/1sin(lim

3/4

x

xx

01

0

)/1cos(

)3/4(

)/1cos()/1(

)3/4(

)/1sin(lim

3/1

2

3/73/4

x

x

xx

x

x

xx

FORMS OF TYPE

Theorem 4.4.2: Suppose that f and g are differentiable functions on an open interval containing x = a, except possibly at x = a, and that

and

If exists or if this limit is then

)(lim xgax

)(lim xfax

/

)](/)([lim // xgxfax

)(

)(lim

)(

)(lim

/

/

xg

xf

xg

xfaxax

EXAMPLE 3

Find the following limits. xx e

xlim

EXAMPLE 3

Find the following limits.

First, we confirm the form of the equation. Next, apply L’Hopital’s Rule.

xx e

xlim

011

lim

xxx ee

x

EXAMPLE 3

Find the following limits.x

xx csc

lnlim0

EXAMPLE 3


First, we confirm the form of the equation. Next, apply L’Hopital’s Rule.

This is the same form as the original, and repeated applications of L’Hopital’s Rule will yield powers of 1/x in the numerator and expressions involving cscx and cotx in the denominator being of no help. We must try another method.

xx

x

x

xx cotcsc

/1

csc

lnlim0

x

xx csc

lnlim0

EXAMPLE 3


We will rewrite the last equation as

x

x

x

xx

x

x

xx

tansin

cotcsc

/1

csc

lnlim0

x

xx csc

lnlim0

0)0)(1(tanlimsin

lim00

xx

xxx

EXAMINE EXPONENTIAL GROWTH

Look at the following graphs. What these graphs show us is that ex grows more rapidly than any integer power of x.

n

x

x x

elim0lim

x

n

x e

x

FORMS OF TYPE

There are several other types of limits that we will encounter. When we look at the form we can make it look like 0/0 or and use L’Hopital’s Rule.

0

0

/

EXAMPLE 4

Evaluate the following limit. xxx

lnlim0

EXAMPLE 4

Evaluate the following limit.

As written, this is of the form

0lnlim0

xxx

xxx

lnlim0

EXAMPLE 4



Lets rewrite this equation as

x

xx /1

lnlim0

0lnlim0

xxx

xxx

lnlim0

EXAMPLE 4




Applying L’Hopital’s Rule

x

xx /1

lnlim0

0lnlim0

xxx

xxx

lnlim0

0/1

/1

/1

lnlim

20

x

x

x

x

xx

EXAMPLE 4

Evaluate the following limit. xxx

2sec)tan1(lim4/

EXAMPLE 4



02sec)tan1(lim4/

xxx

xxx

2sec)tan1(lim4/

EXAMPLE 4



Lets rewrite this equation asx

xx 2cos

tan1lim

4/

02sec)tan1(lim4/

xxx

xxx

2sec)tan1(lim4/

EXAMPLE 4




Applying L’Hopital’s Rule

x

xx 2cos

tan1lim

4/

02sec)tan1(lim4/

xxx

xxx

2sec)tan1(lim4/

12

2

2sin2

sec

2cos

tan1lim

2

4/

x

x

x

xx

FORMS OF TYPE

There is another form we will look at. This is a limit of the form We will again try to manipulate this equation to match a form we like.

.

EXAMPLE 5


xxx sin

11lim0

EXAMPLE 5


This is of the form

xxx sin

11lim0

EXAMPLE 5


This is of the form

Lets rewrite the equation as

xxx sin

11lim0

xx

xx

xxx sin

sin

sin

11lim0

EXAMPLE 5


This is of the form

Lets rewrite the equation as

This is of the form 0/0, so we will use L’Hopital’s Rule.

xxx sin

11lim0

xx

xx

xxx sin

sin

sin

11lim0

02

0

cos2sin

sin

0

0

sincos

1cos

sin

sinlim0

xxx

x

xxx

x

xx

xxx

FORMS OF TYPE

Limits of the form give rise to other indeterminate forms of the types:

)()(lim xgxf

1,,0 00

1,,0 00

FORMS OF TYPE

Limits of the form give rise to other indeterminate forms of the types:

Equations of this type can sometimes be evaluated by first introducing a dependent variable

and then finding the limit of lny.

)()(lim xgxf

1,,0 00

1,,0 00

)()( xgxfy

EXAMPLE 6

Show that ex x

x

/1

0)1(lim

EXAMPLE 6

Show that

This is of the form so we will introduce a dependent variable and solve.

ex x

x

/1

0)1(lim

1

EXAMPLE 6

Show that

This is of the form so we will introduce a dependent variable and solve.

ex x

x

/1

0)1(lim

1

xxy /1)1(

xxy /1)1ln(ln

)1ln()/1(ln xxy

x

xy

)1ln(ln

EXAMPLE 6

Show that ex x

x

/1

0)1(lim

x

xy

)1ln(ln

0

0)1ln(limlnlim

00

x

xy

xx

EXAMPLE 6

Show that ex x

x

/1

0)1(lim

x

xy

)1ln(ln

0

0)1ln(limlnlim

00

x

xy

xx

11

1

1

)1/(1)1ln(lim

0

x

x

x

xx

EXAMPLE 6

Show that

This implies that as .

ex x

x

/1

0)1(lim

ey

11

1

1

)1/(1)1ln(lim

0

x

x

x

xx

1lnlim0

yx

0x

Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule

Indeterminate Forms


Indeterminate Forms


Indeterminate Forms


Indeterminate Forms


Other Indeterminate Forms







Find a common denominator


Other Indeterminate Forms – Indeterminate Powers






Documents

LESSON 64 L’HOPITAL’S RULE HL MATH - CALCULUS Objective: To use L’Hopital’s Rule to solve problems involving limits