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
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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
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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
Lets look at a general example.
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
Lets look at a general example.
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
Find the following limit.
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
Find the following limit.
First, we confirm that this of the form 0/0. Next, we use L’Hopital’s Rule to find the limit.
x
xx cos
sin1lim
2/
01
0
sin
cos
cos
sin1lim
2/
x
x
x
xx
EXAMPLE 2
Find the following limit. 30
1lim
x
ex
x
EXAMPLE 2
Find the following limit.
First, we confirm that this of the form 0/0. Next, we use L’Hopital’s Rule to find the limit.
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
Find the following limit. 20
tanlim
x
xx
EXAMPLE 2
Find the following limit.
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
Find the following limit. 20
cos1lim
x
xx
EXAMPLE 2
Find the following limit.
First, we confirm that this of the form 0/0. Next we use L’Hopital’s Rule to find the answer.
We apply L’Hopital’s Rule again.
0
0
2
sincos1lim
20
x
x
x
xx
20
cos1lim
x
xx
EXAMPLE 2
Find the following limit.
First, we confirm that this of the form 0/0. Next we use L’Hopital’s Rule to find the answer.
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
Find the following limit.
First, we confirm that this of the form 0/0. Next we use L’Hopital’s Rule to find the answer.
)/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
Find the following limit.
First, we confirm that this of the form 0/0. Next we use L’Hopital’s Rule to find the answer.
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
Find the following limits.
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
Find the following limits.
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
Evaluate the following limit.
As written, this is of the form
Lets rewrite this equation as
x
xx /1
lnlim0
0lnlim0
xxx
xxx
lnlim0
EXAMPLE 4
Evaluate the following limit.
As written, this is of the form
Lets rewrite this equation as
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
Evaluate the following limit.
As written, this is of the form
02sec)tan1(lim4/
xxx
xxx
2sec)tan1(lim4/
EXAMPLE 4
Evaluate the following limit.
As written, this is of the form
Lets rewrite this equation asx
xx 2cos
tan1lim
4/
02sec)tan1(lim4/
xxx
xxx
2sec)tan1(lim4/
EXAMPLE 4
Evaluate the following limit.
As written, this is of the form
Lets rewrite this equation as
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
Find the following limit.
xxx sin
11lim0
EXAMPLE 5
Find the following limit.
This is of the form
xxx sin
11lim0
EXAMPLE 5
Find the following limit.
This is of the form
Lets rewrite the equation as
xxx sin
11lim0
xx
xx
xxx sin
sin
sin
11lim0
EXAMPLE 5
Find the following limit.
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
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms
Find a common denominator
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms – Indeterminate Powers
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms – Indeterminate Powers
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Sec 4.5 – Indeterminate Forms and L’Hopital’s Rule
Other Indeterminate Forms – Indeterminate Powers