Solve this naturally

Look here! $y = \lim_{x \rightarrow 0^+} (\sin(x))^{\frac{1}{\ln (x)}}$ $\ln(y) = \lim_{x \rightarrow 0^+} \frac{\ln(\sin(x))}{\ln(x)}$ With L'hopital (since it is $\infty/\infty$ ), then we have

$\ln(y) = \lim_{x \rightarrow 0^+} \frac{\ln(\sin(x))}{\ln(x)}$ $\ln(y) = \lim_{x \rightarrow 0^+} \frac{\cot (x))}{\frac{1}{x}}$ $\ln(y) = \lim_{x \rightarrow 0^+} \frac{x}{\tan (x)}$

One more time L'hopital ( $0/0$

$\ln(y) = \lim_{x \rightarrow 0^+} \frac{1}{\sec ^2 (x)}$ Evaluate the limit, we have $\ln(y) = 1$ Exponent it, yielding to $y=e$

It is not necessary to go as far as other solutions. This result is a consequence of the fact that $x \sim \sin x$ around 0. Indeed: $\lim_{x \to 0^+} \left(\frac{x}{\sin x} \right)^{\frac{1}{\log x}} = 1^{0} = 1$ $\lim_{x \to 0^+} (\sin x)^{\frac{1}{\log x}} = \lim_{x \to 0^+} x^{\frac{1}{\log x}} = \lim_{x \to 0^+} e^{\log x \frac{1}{\log x}} = \lim_{x \to 0^+} e = e$

apologies for that first image.......