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SOLUTION We rst notice that both the numerator and the denominator tend to + as x + Thus the quotient is indeterminate at + of the form + / + Therefore the new version of l H pital s Rule applies and our limit equals 4x 3 x + ex Again the numerator and denominator tend to + as x + , so we once more apply l H pital s Rule The limit equals lim 12x 2 = 0 x + ex We must apply l H pital s Rule two more times We rst obtain lim 24x x + ex lim
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CHAPTER 5 Indeterminate Forms
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and then 24 x + ex lim We conclude that x4 = 0 x + ex ex You Try It: Evaluate the limit limx + x ln x You Try It: Evaluate the limit limx x 4 ex lim EXAMPLE 55
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Evaluate the limit
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sin(2/x) sin(5/x)
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SOLUTION We note that both numerator and denominator tend to 0, so the quotient is indeterminate at of the form 0/0 We may therefore apply l H pital s Rule Our limit equals ( 2/x 2 ) cos(2/x) x ( 5/x 2 ) cos(5/x) lim This in turn simpli es to 2 2 cos(2/x) = x 5 cos(5/x) 5 lim l H pital s Rule also applies to one-sided limits Here is an example EXAMPLE 56
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Evaluate the limit
x 0+
x
SOLUTION Both numerator and denominator tend to zero so the quotient is indeterminate at 0 of the form 0/0 We may apply l H pital s Rule; differentiating numerator and denominator, we nd that the limit equals [cos x] (1/2)x 1/2 lim = lim cos x 1/2 + + (1/2)x x 0 x 0 = 1 You Try It: How can we apply l H pital s Rule to evaluate limx 0+ x ln x
Indeterminate Forms
Other Indeterminate Forms
521 INTRODUCTION
By using some algebraic manipulations, we can reduce a variety of indeterminate limits to expressions which can be treated by l H pital s Rule We explore some of these techniques in this section
WRITING A PRODUCT AS A QUOTIENT
The technique of the rst example is a simple one, but it is used frequently EXAMPLE 57
Evaluate the limit
x
lim x 2 e3x
SOLUTION Notice that x 2 + while e3x 0 So the limit is indeterminate of the form 0 We rewrite the limit as
x e 3x
Now both numerator and denominator tend to in nity and we may apply l H pital s Rule The result is that the limit equals 2x x 3e 3x lim
Again the numerator and denominator both tend to in nity so we apply l H pital s Rule to obtain:
x
It is clear that the limit of this last expression is zero We conclude that
x
You Try It: Evaluate the limit limx + e
THE USE OF THE LOGARITHM
The natural logarithm can be used to reduce an expression involving exponentials to one involving a product or a quotient
AM FL Y
lim x2 lim 2 9e 3x lim x e3x = 0
x
CHAPTER 5 Indeterminate Forms
EXAMPLE 58
Evaluate the limit
x 0+
lim x x
SOLUTION We study the limit of f (x) = x x by considering ln f (x) = x ln x We rewrite this as ln x lim ln f (x) = lim x 0+ 1/x x 0+ Both numerator and denominator tend to , so the quotient is indeterminate of the form / Thus l H pital s Rule applies The limit equals
x 0+
1/x = lim x = 0 1/x 2 x 0+
Now the only way that ln f (x) can tend to zero is if f (x) = x x tends to 1 We conclude that
x 0+
lim x x = 1
EXAMPLE 59
Evaluate the limit
x 0
lim (1 + x 2 )ln |x |
SOLUTION Let f (x) = (1 + x 2 )ln |x| and consider ln f (x) = ln |x| ln(1 + x 2 ) This expression is indeterminate of the form 0 We rewrite it as ln(1 + x 2 ) , x 0 1/ ln |x| lim so that both the numerator and denominator tend to 0 So l H pital s Rule applies and we have lim ln f (x) = lim 2x/(1 + x 2 ) 1/[x ln2 (|x|)] = lim
2x 2 ln2 (|x|) (1 + x 2 )
The numerator tends to 0 (see Example 53) and the denominator tends to 1 Thus
lim ln f (x) = 0
Indeterminate Forms
But the only way that ln f (x) can tend to zero is if f (x) tends to 1 We conclude that
lim (1 + x 2 )ln |x| = 1
You Try It: Evaluate the limit limx 0+ (1/x)x You Try It: Evaluate the limit limx 0+ (1 + x)1/x In fact this limit gives an important way to de ne Euler s constant e (see Sections 19 and 623)