Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 37

Answer

$$1$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} \cr & {\text{Evaluating}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = {0^{\sin 0}} = {0^0} \cr & {\text{This limit has the form }}{0^0}.{\text{ }} \cr & {x^{\sin x}} = {e^{\sin x\ln x}},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\sin x\ln x}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x}} \cr & {\text{The first step is to evaluate }} \cr & L = \mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sin x}}}} = \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sin x}}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{1}{x}}}{{ - \frac{{\cos x}}{{\sin x}}}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{x\cos x}} = \frac{0}{0} \cr & {\text{Using the L'Hopital's rule}} \cr & - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{x\cos x}} = - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos x}}{{x\sin x + \cos x}} = \frac{0}{{0 + 1}} = 0 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\sin x\ln x}} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \sin x\ln x}} = {e^0} = 1 \cr & \mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin x}} = 1 \cr} $$
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