applications of differentiation 4. 4.4 indeterminate forms and l’hospital’s rule applications of...
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APPLICATIONS OF DIFFERENTIATIONAPPLICATIONS OF DIFFERENTIATION
4
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4.4Indeterminate Forms
and L’Hospital’s Rule
APPLICATIONS OF DIFFERENTIATION
In this section, we will learn:
How to evaluate functions whose
values cannot be found at certain points.
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Suppose we are trying to analyze
the behavior of the function
Although F is not defined when x = 1,
we need to know how F behaves near 1.
ln( )
1
xF x
x
INDETERMINATE FORMS
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In particular, we would like to know
the value of the limit
1
lnlim
1x
x
x
Expression 1INDETERMINATE FORMS
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In computing this limit, we can’t apply
Law 5 of limits (Section 2.3) because
the limit of the denominator is 0.
In fact, although the limit in Expression 1 exists, its value is not obvious because both numerator and denominator approach 0 and is not defined.0
0
INDETERMINATE FORMS
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INDETERMINATE FORM —TYPE 0/0
In general, if we have a limit of the form
where both f(x) → 0 and g(x) → 0 as x → a,
then this limit may or may not exist.
It is called an indeterminate form of type .
We met some limits of this type in Chapter 2.
( )lim
( )x a
f x
g x
0
0
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INDETERMINATE FORMS
For rational functions, we can cancel
common factors:
2
21 1
1
( 1)lim lim
( 1)( 1)1
1lim
1 2
x x
x
x x x x
x xx
x
x
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INDETERMINATE FORMS
We used a geometric argument
to show that:
0
sinlim 1x
x
x
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INDETERMINATE FORMS
However, these methods do not work
for limits such as Expression 1.
Hence, in this section, we introduce a systematic method, known as l’Hospital’s Rule, for the evaluation of indeterminate forms.
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INDETERMINATE FORMS
Another situation in which a limit is
not obvious occurs when we look for
a horizontal asymptote of F and need
to evaluate the limit lnlim
1x
x
x
Expression 2
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INDETERMINATE FORMS
It isn’t obvious how to evaluate this limit
because both numerator and denominator
become large as x → ∞.
There is a struggle between the two. If the numerator wins, the limit will be ∞. If the denominator wins, the answer will be 0. Alternatively, there may be some compromise—
the answer may be some finite positive number.
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INDETERMINATE FORM —TYPE ∞/∞
In general, if we have a limit of the form
where both f(x) → ∞ (or -∞) and g(x) → ∞
(or -∞), then the limit may or may not exist.
It is called an indeterminate form of type ∞/∞.
( )lim
( )x a
f x
g x
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INDETERMINATE FORMS
We saw in Section 2.6 that this type of limit
can be evaluated for certain functions—
including rational functions—by dividing the
numerator and denominator by the highest
power of x that occurs in the denominator.
For instance,2 2
2
2
11
1 1 0 1lim lim
1 2 0 22 1 2x x
x xx
x
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INDETERMINATE FORMS
This method, though, does not work
for limits such as Expression 2.
However, L’Hospital’s Rule also applies to this type of indeterminate form.
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L’HOSPITAL’S RULE
Suppose f and g are differentiable and
g’(x) ≠ 0 on an open interval I that contains a
(except possibly at a).
Suppose
or that
In other words, we have an indeterminate form of type or ∞/∞.
lim ( ) 0 and lim ( ) 0x a x a
f x g x
0
0
lim ( ) and lim ( )x a x a
f x g x
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L’HOSPITAL’S RULE
Then,
if the limit on the right exists
(or is ∞ or - ∞).
( ) '( )lim lim
( ) '( )x a x a
f x f x
g x g x
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NOTE 1
L’Hospital’s Rule says that the limit of
a quotient of functions is equal to the limit of
the quotient of their derivatives—provided that
the given conditions are satisfied.
It is especially important to verify the conditions regarding the limits of f and g before using the rule.
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NOTE 2
The rule is also valid for one-sided limits
and for limits at infinity or negative infinity.
That is, ‘x → a’ can be replaced by any of the symbols x → a+, x → a-, x → ∞, or x → - ∞.
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NOTE 3
For the special case in which
f(a) = g(a) = 0, f’ and g’ are continuous,
and g’(a) ≠ 0, it is easy to see why
the rule is true.
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NOTE 3
In fact, using the alternative form of
the definition of a derivative, we have:
( ) ( ) ( ) ( )lim
'( ) '( )lim lim
( ) ( ) ( ) ( )'( ) '( ) lim
( ) ( )lim
( ) ( )
( )lim
( )
x a
x a x a
x a
x a
x a
f x f a f x f af x f a x a x a
g x g a g x g ag x g ax a x a
f x f a
g x g a
f x
g x
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NOTE 3
It is more difficult to prove
the general version of l’Hospital’s
Rule.
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L’HOSPITAL’S RULE
Find
and
Thus, we can apply l’Hospital’s Rule:
1
lnlim
1x
x
x
1 1lim ln ln1 0 lim( 1) 0x x
x x
1 1 1 1
(ln )ln 1/ 1
lim lim lim lim 11 1( 1)
x x x x
dx
x xdxdx xxdx
Example 1
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L’HOSPITAL’S RULE
Calculate
We have and
So, l’Hospital’s Rule gives:
2lim
x
x
e
x
2lim limx
x xe x
22
( )lim lim lim
2( )
xx x
x x x
dee edx
dx xxdx
Example 2
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L’HOSPITAL’S RULE
As ex → ∞ and 2x → ∞ as x → ∞, the limit
on the right side is also indeterminate.
However, a second application of l’Hospital’s
Rule gives: 2
lim lim lim2 2
x x x
x x x
e e e
x x
Example 2
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L’HOSPITAL’S RULE
Calculate
As ln x → ∞ and as x → ∞, l’Hospital’s Rule applies:
Notice that the limit on the right side is now indeterminate of type .
3
lnlimx
x
x
3 x
2/3133
ln 1/lim limx x
x x
xx
0
0
Example 3
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L’HOSPITAL’S RULE
However, instead of applying the rule a second time as we did in Example 2, we simplify the expression and see that a second application is unnecessary:
2 / 313 33
ln 1/ 3lim lim lim 0x x x
x x
xx x
Example 3
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L’HOSPITAL’S RULE
Find
Noting that both tan x – x → 0 and x3 → 0 as x → 0, we use l’Hospital’s Rule:
30
tanlimx
x x
x
2
3 20 0
tan sec 1lim lim
3x x
x x x
x x
Example 4
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L’HOSPITAL’S RULE
As the limit on the right side is still indeterminate of type , we apply the rule again:
Example 4
2 2
20 0
sec 1 2sec tanlim lim
63x x
x x x
xx
0
0
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L’HOSPITAL’S RULE
Since , we simplify the calculation by writing:
2
0limsec 1x
x
22
0 0 0
0
2sec tan 1 tanlim limsec lim
6 31 tan
lim3
x x x
x
x x xx
x xx
x
Example 4
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L’HOSPITAL’S RULE
We can evaluate this last limit either by using l’Hospital’s Rule a third time or by writing tan x as (sin x)/(cos x) and making use of our knowledge of trigonometric limits.
Example 4
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L’HOSPITAL’S RULE
Putting together all the steps,
we get:2
3 20 0
2
0
2
0 0
tan sec 1lim lim
3
2sec tanlim
6
1 tan 1 sec 1lim lim
3 3 1 3
x x
x
x x
x x x
x x
x x
x
x x
x
Example 4
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L’HOSPITAL’S RULE
Find
If we blindly attempted to use l-Hospital’s rule, we would get:
sinlim
1 cosx
x
x
sin coslim lim
1 cos sinx x
x x
x x
Example 5
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This is wrong.
Although the numerator sin x → 0 as x → π -, notice that the denominator (1 - cos x) does not approach 0.
So, the rule can’t be applied here.
L’HOSPITAL’S RULE Example 5
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L’HOSPITAL’S RULE
The required limit is, in fact, easy to find
because the function is continuous at π
and the denominator is nonzero there:
sin sin 0lim 0
1 cos 1 cos 1 ( 1)x
x
x
Example 5
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L’HOSPITAL’S RULE
The example shows what can go wrong
if you use the rule without thinking.
Other limits can be found using the rule, but are more easily found by other methods.
See Examples 3 and 5 in Section 2.3, Example 3 in Section 2.6, and the discussion at the beginning of the section.
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L’HOSPITAL’S RULE
So, when evaluating any limit,
you should consider other methods
before using l’Hospital’s Rule.
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INDETERMINATE PRODUCTS
If and (or -∞),
then it isn’t clear what the value
of , if any, will be.
lim ( ) 0x a
f x
lim ( )
x ag x
lim ( ) ( )x a
f x g x
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INDETERMINATE PRODUCTS
There is a struggle between f and g.
If f wins, the answer will be 0.
If g wins, the answer will be ∞ (or -∞).
Alternatively, there may be a compromise where the answer is a finite nonzero number.
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This kind of limit is called an indeterminate
form of type 0 . ∞.
We can deal with it by writing the product fg as a quotient:
This converts the given limit into an indeterminate form of type or ∞/∞, so that we can use l’Hospital’s Rule.
or1/ 1/
f gfg fg
g f
0
0
INDETERMINATE FORM—TYPE 0 . ∞
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INDETERMINATE PRODUCTS
Evaluate
The given limit is indeterminate because, as x → 0+, the first factor (x) approaches 0, whereas the second factor (ln x) approaches -∞.
0lim lnx
x x
Example 6
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INDETERMINATE PRODUCTS
Writing x = 1/(1/x), we have 1/x → ∞ as x → 0+.
So, l’Hospital’s Rule gives:
Example 6
20 0 0
0
ln 1/lim ln lim lim
1/ 1/lim ( ) 0
x x x
x
x xx x
x xx
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In solving the example, another possible
option would have been to write:
This gives an indeterminate form of the type 0/0.
However, if we apply l’Hospital’s Rule, we get a more complicated expression than the one we started with.
0 0lim ln lim
1/ lnx x
xx x
x
INDETERMINATE PRODUCTS Note
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INDETERMINATE PRODUCTS
In general, when we rewrite an
indeterminate product, we try to choose
the option that leads to the simpler limit.
Note
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If and , then
the limit
is called an indeterminate form
of type ∞ - ∞.
lim ( )x a
f x
lim ( )x ag x
lim[ ( ) ( )]x a
f x g x
INDETERMINATE FORM—TYPE ∞ - ∞
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INDETERMINATE DIFFERENCES
Again, there is a contest between f and g.
Will the answer be ∞ (f wins)?
Will it be - ∞ (g wins)?
Will they compromise on a finite number?
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INDETERMINATE DIFFERENCES
To find out, we try to convert the difference
into a quotient (for instance, by using a
common denominator, rationalization, or
factoring out a common factor) so that
we have an indeterminate form of type
or ∞/∞.
0
0
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INDETERMINATE DIFFERENCES
Compute
First, notice that sec x → ∞ and tan x → ∞ as x → (π/2)
-.
So, the limit is indeterminate.
( / 2)lim (sec tan )
xx x
Example 7
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INDETERMINATE DIFFERENCES
Here, we use a common denominator:
Note that the use of l’Hospital’s Rule is justified because 1 – sin x → 0 and cos x → 0 as x → (π/2)
-.
( / 2) ( / 2)
( / 2)
( / 2)
1 sinlim (sec tan ) lim
cos cos
1 sinlim
coscos
lim 0sin
x x
x
x
xx x
x x
x
xx
x
Example 7
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INDETERMINATE POWERS
Several indeterminate forms arise from
the limit( )lim[ ( )]g x
x af x
0
0
1. lim ( ) 0 and lim ( ) 0 type0
2. lim ( ) and lim ( ) 0 type
3. lim ( ) 1 and lim ( ) type1
x a x a
x a x a
x a x a
f x g x
f x g x
f x g x
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INDETERMINATE POWERS
Each of these three cases can be treated
in either of two ways.
Taking the natural logarithm:
Writing the function as an exponential:
( )Let [ ( )] , then ln ( ) ln ( )g xy f x y g x f x
( ) ( ) ln ( )[ ( )]g x g x f xf x e
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INDETERMINATE POWERS
Recall that both these methods were used
in differentiating such functions.
In either method, we are led to the indeterminate product g(x) ln f(x), which is of type 0 . ∞.
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INDETERMINATE POWERS
Calculate
First, notice that, as x → 0+, we have 1 + sin 4x → 1 and cot x → ∞.
So, the given limit is indeterminate.
cot
0lim (1 sin 4 ) x
xx
Example 8
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INDETERMINATE POWERS
Let y = (1 + sin 4x)cot x
Then, ln y = ln[(1 + sin 4x)cot x]
= cot x ln(1 + sin 4x)
Example 8
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INDETERMINATE POWERS
So, l’Hospital’s Rule gives:
0 0
20
ln(1 sin 4 )lim ln lim
tan4cos 4
1 sin 4lim 4sec
x x
x
xy
xx
xx
Example 8
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INDETERMINATE POWERS
So far, we have computed the limit of ln y.
However, what we want is the limit of y.
To find this, we use the fact that y = eln y:
cot
0 0
ln 4
0
lim(1 sin 4 ) lim
lim
x
x x
y
x
x y
e e
Example 8
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INDETERMINATE POWERS
Find
Notice that this limit is indeterminate since 0
x = 0 for any x > 0 but x0 = 1
for any x ≠ 0.
0lim x
xx
Example 9
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INDETERMINATE POWERS
We could proceed as in Example 8 or by
writing the function as an exponential:
xx = (eln x)x = ex ln x
In Example 6, we used l’Hospital’s Rule to show that
Therefore,
0lim ln 0x
x x
ln 0
0 0lim lim 1x x x
x xx e e
Example 9