TITLE: Is my method for computing $\lim_{x \to 0^+} \left( \dfrac{\sin x}{x} \right)^{1/x^2} = e^{-1/6}$ valid?
QUESTION [7 upvotes]: I know that if $\lim_{x \to a}g(x)$ exists and $f(x)$ is continuous at $\lim_{x \to a}g(x)$, then we can interchange the limit with $f(x)$. That is,
$$
\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x)).
$$
So onto the limit in question:
I'd like to show that
$$
\lim_{x \to 0^+} \left( \dfrac{\sin x}{x} \right)^{1/x^2} = e^{-1/6}.
$$
I start off by letting
$$
L = \lim_{x \to 0^+}\left(\dfrac{\sin x}{x} \right)^{1/x^2}.
$$
Next, I apply the natural logarithm to both sides of the equation to give $$
\ln L = \ln\left(\lim_{x \to 0^+}\left(\dfrac{\sin x}{x} \right)^{1/x^2}\right).
$$
But now, I would like to interchange the natural log with the limit on the right side of the equation. However, this seems not justified because I don't know, a priori, that $\lim_{x \to 0^+} \left( \dfrac{\sin x}{x} \right)^{1/x^2}$ exists and if it does exist, that the natural log is even defined at the limit (what if it's $0$ or negative?).
Thus, my method of interchanging the natural log with the limit seems circular. I need to assume the limit exists and is neither $0$ nor negative before proceeding with my computation.
So here are my questions: Is it actually justified to interchange the natural log with the limit in this type of computation? Is it commonplace and accepted to assume the limit exists when going about these types of computations? If not, can someone propsose an alternative method for computing this limit without using the epsilon-delta definition?

REPLY [3 votes]: You can assume
$$L>0\implies\left(\lim_{x\to0^+}f(x)=L\iff \lim_{x\to0^+}\log(f(x))=\log(L)\right)$$ by continuity and monotonicity of the logarithm (in its domain).
Now you can evaluate
$$M=\lim_{x\to 0^+}\frac{\log\left(\dfrac{\sin x}{x}\right)}{x^2}$$
and conclude for $L$. (Note that it works as soon as $M$ exists.)