Chapter 4 | Applications of Derivatives
457
b. As x →1, the numerator sin( πx )→0 and the denominator ln( x )→0. Therefore, we can apply L’Hôpital’s rule. We obtain lim x →1 sin( πx ) ln x = lim x →1 π cos( πx ) 1/ x = lim x →1 ( πx )cos( πx ) = ( π ·1)(−1) =− π . c. As x →∞, the numerator e 1/ x −1→0 and the denominator ⎛ ⎝ 1 x ⎞ ⎠ →0. Therefore, we can apply L’Hôpital’s rule. We obtain
e 1/ x ⎛
⎞ ⎠
⎝ −1 x 2
e 1/ x −1 1 x
e 1/ x = e 0 =1.
⎛ ⎝ −1 x 2
⎞ ⎠
lim x →∞
= lim x →∞
= lim x →∞
d. As x →0, both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain lim x →0 sin x − x x 2 = lim x →0 cos x −1 2 x . Since the numerator and denominator of this new quotient both approach zero as x →0, we apply L’Hôpital’s rule again. In doing so, we see that lim x →0 cos x −1 2 x = lim x →0 −sin x 2 =0.
Therefore, we conclude that
sin x − x x 2
lim x →0
=0.
x tan x .
4.37
Evaluate lim x →0
We can also use L’Hôpital’s rule to evaluate limits of quotients f ( x ) g ( x ) inwhich f ( x )→±∞ and g ( x )→±∞. Limits of this form are classified as indeterminate forms of type ∞/∞. Again, note that we are not actually dividing ∞ by ∞. Since ∞ is not a real number, that is impossible; rather, ∞/∞. is used to represent a quotient of limits, each of which is ∞ or −∞. Theorem 4.13: L’Hôpital’s Rule (∞/∞ Case) Suppose f and g are differentiable functions over an open interval containing a , except possibly at a . Suppose lim x → a f ( x ) =∞ (or −∞) and lim x → a g ( x ) =∞ (or −∞). Then, lim x → a f ( x ) g ( x ) = lim x → a f ′( x ) g ′( x ) , assuming the limit on the right exists or is ∞ or −∞. This result also holds if the limit is infinite, if a =∞ or
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