How to approach it?

Assuming that the limit is as $x \rightarrow 0^{+}$ , then we first note that the limit is of the indeterminate form $0^{0}$ .

So we look instead at

$\ln(L) = \ln(\lim_{x \rightarrow 0^{+}} (\sin(x))^{\frac{1}{\ln(x)}}) = \lim_{x \rightarrow 0^{+}} (\ln((\sin(x))^{\frac{1}{\ln(x)}}) = \lim_{x \rightarrow 0^{+}} \dfrac{\ln(\sin(x))}{\ln(x)}$ .

Now apply L'Hopital's rule to this last expression to find that

$\ln(L) = \lim_{x \rightarrow 0^{+}} \dfrac{\frac{\cos(x)}{\sin(x)}}{\frac{1}{x}} = \lim_{x \rightarrow 0^{+}} \dfrac{\cos(x)}{\frac{\sin(x)}{x}} = \dfrac{1}{1} = 1$ ,

where I made use of the fact that $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$ .

So since $\ln(L)$ goes to $1$ as $x \rightarrow 0^{+}$ , we conclude that $L$ goes to $e^{1} = e = 2.71828.....$ . Thus $\lfloor L \rfloor = \boxed{2}$ .