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

$${\text{True}}$$

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