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Chapter 4 | Applications of Derivatives
the indeterminate forms in such a way that we arrive at the indeterminate form 0 0
or ∞/∞.
Indeterminate Form of Type 0·∞ Suppose we want to evaluate lim x → a ⎛
⎝ f ( x ) · g ( x ) ⎞ ⎠ , where f ( x )→0 and g ( x )→∞ (or −∞) as x → a . Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation 0·∞ to denote the form that arises in this situation. The expression 0·∞ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product f ( x ) g ( x ) as x → a . For example, let n be a positive integer and consider f ( x ) = 1 ( x n +1) and g ( x ) =3 x 2 . As x →∞, f ( x )→0 and g ( x )→∞. However, the limit as x →∞ of f ( x ) g ( x ) = 3 x 2 ( x n +1) varies, depending on n . If n =2, then lim x →∞ f ( x ) g ( x ) =3. If n =1, then lim x →∞ f ( x ) g ( x ) =∞. If n =3, then lim x →∞ f ( x ) g ( x ) =0. Herewe consider another limit involving the indeterminate form 0·∞ and show how to rewrite the function as a quotient to use L’Hôpital’s rule. Example 4.41 Indeterminate Form of Type 0·∞
Evaluate lim x →0 +
x ln x .
Solution First, rewrite the function x ln x as a quotient to apply L’Hôpital’s rule. If we write x ln x = ln x 1/ x , we see that ln x →−∞ as x →0 + and 1 x →∞ as x →0
+ . Therefore, we can apply L’Hôpital’s rule and
obtain
d dx (ln x ) d dx (1/ x )
1/ x -1/ x 2
ln x 1/ x = lim
(− x ) =0.
lim x →0 +
= lim
= lim
x →0 +
x →0 +
x →0 +
We conclude that
lim x →0 +
x ln x =0.
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