The value of lim _{x rightarrow 0^+} ((sin x) | vedclass

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(C) Let $L = \lim _{x \rightarrow 0^+} ((\sin x)^{\frac{1}{x}} + (\frac{1}{x})^{\sin x})$.First,consider $L_1 = \lim _{x \rightarrow 0^+} (\sin x)^{\f

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(C) Let $L = \lim _{x ightarrow 0^+} ((\sin x)^{\frac{1}{x}} + (\frac{1}{x})^{\sin x})$.First,consider $L_1 = \lim _{x ightarrow 0^+} (\sin x)^{\frac{1}{x}}$. As $x ightarrow 0^+$,$\sin x ightarrow 0^+$ and $\frac{1}{x} ightarrow \infty$. Since $0$ raised to any positive power is $0$,$L_1 = 0$.Now,consider $L_2 = \lim _{x ightarrow 0^+} (\frac{1}{x})^{\sin x}$.Taking the natural logarithm on both sides: $\ln L_2 = \lim _{x ightarrow 0^+} \sin x \ln(\frac{1}{x}) = \lim _{x ightarrow 0^+} -\sin x \ln x$.This is an indeterminate form of type $0 	imes \infty$. We rewrite it as: $\ln L_2 = -\lim _{x ightarrow 0^+} \frac{\ln x}{\csc x}$.Applying $L'H\hat{o}pital's$ Rule: $\ln L_2 = -\lim _{x ightarrow 0^+} \frac{1/x}{-\csc x \cot x} = \lim _{x ightarrow 0^+} \frac{\sin x 	an x}{x} = \lim _{x ightarrow 0^+} (\frac{\sin x}{x}) \cdot 	an x$.Since $\lim _{x ightarrow 0^+} \frac{\sin x}{x} = 1$ and $\lim _{x ightarrow 0^+} 	an x = 0$,we have $\ln L_2 = 1 	imes 0 = 0$.Thus,$L_2 = e^0 = 1$.Therefore,$L = L_1 + L_2 = 0 + 1 = 1$.

The value of $\lim _{x \rightarrow 0^+} ((\sin x)^{\frac{1}{x}} + (\frac{1}{x})^{\sin x})$,where $x > 0$,is

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