Chapter 4 | Applications of Derivatives
463
0·∞, and we can use the techniques discussed earlier to rewrite the expression g ( x )ln ⎛ ⎝ f ( x ) ⎞
⎠ in a form so that we can
g ( x )ln ⎛
⎞ ⎠ = L , where L may be ∞ or −∞. Then lim x → a ⎡ ⎣ ln( y ) ⎤ ⎦ = L .
⎝ f ( x )
apply L’Hôpital’s rule. Suppose lim x → a
Since the natural logarithm function is continuous, we conclude that ln ⎛ ⎝ lim x → a y ⎞ ⎠ = L , which gives us lim x → a y = lim x → a
f ( x ) g ( x ) = e L .
Example 4.43 Indeterminate Form of Type ∞ 0
x 1/ x .
Evaluate lim x →∞
Solution Let y = x 1/ x . Then,
⎛ ⎝ x 1/ x
⎞ ⎠ = 1 x ln x = ln x x .
ln
ln x x . Applying L’Hôpital’s rule, we obtain lim x →∞ ln y = lim x →∞ ln x x = lim x →∞ 1/ x
We need to evaluate lim x →∞
1 =0.
Therefore, lim x
→∞ ln y =0. Since the natural logarithm function is continuous, we conclude that ln ⎛ ⎝ lim x →∞ y ⎞ ⎠ =0,
which leads to
ln x
0 =1.
lim x →∞
y = lim x
x = e
→∞
Hence,
x 1/ x =1.
lim x →∞
x 1/ln( x ) .
4.42
Evaluate lim x →∞
Example 4.44 Indeterminate Form of Type 0 0
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