Is my method for computing $\lim_{x \to 0^+} \left( \dfrac{\sin x}{x} \right)^{1/x^2} = e^{-1/6}$ valid? 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?
 A: Formally you can compute $V=\lim_{x\to 0^+}\ln\left(\left(x^{-1}\sin x\right)^{x^{-2}}\right)$ and then say that $L=e^V$ with the theorem you've mentioned, applied to the function $e^x$ instead.
More colloquially, since $\ln:(0,\infty)\to \Bbb R$ is a (strictly increasing) homeomorphism, you can say that for a positive function $f$ the quantity $\lim_{x\to a}f(x)$ exists as an extended real number if and only if $\lim_{x\to a}\ln f(x)$ exists, and if so they are related in the obvious way.
A: 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.)
A: I will use the follow identity
$$\left( \dfrac{\sin x}{x} \right)^{1/x^2}=e^{\frac{1}{x^2}\ln \left(\frac{\sin \left(x\right)}{x}\right)}=e^{\left(\frac{\frac{1}{x^2}}{\frac{1}{\ln \left(\frac{\sin \left(x\right)}{x}\right)}}\right)}=e^{\left(\frac{\ln \left(\frac{\sin \left(x\right)}{x}\right)}{x^2}\right)}$$
Hence using de L'Hopital rule for $x\to 0^+$,
$$\lim _{x\to 0^+}\left(\frac{\frac{x\cos \left(x\right)-\sin \left(x\right)}{x\sin \left(x\right)}}{2x}\right)=\lim _{x\to 0^+}\left(\frac{x\cos \left(x\right)-\sin \left(x\right)}{2x^2\sin \left(x\right)}\right)$$
Using again the same rule,
$$=\lim_{x\to 0^+}\left(\frac{-x\sin \left(x\right)}{2\left(2x\sin \left(x\right)+\cos \left(x\right)x^2\right)}\right)=\lim _{x\to 0^+}\left(-\frac{\sin \left(x\right)}{2\left(x\cos \left(x\right)+2\sin \left(x\right)\right)}\right)$$
Apply again the same rule,
$$=\lim _{x\to 0^+}\left(\frac{-\cos \left(x\right)}{2\left(-x\sin \left(x\right)+3\cos \left(x\right)\right)}\right)=-\frac 16$$
Definitively the limit it is $e^{-\frac 16}$.
A: Since $\;\sin x=x-\dfrac{x^3}6+o\left(x^3\right)\;,\;$ it follows that
$\lim\limits_{x\to0}\dfrac{\sin -x}{x^3}=\lim\limits_{x\to0}\dfrac{-\dfrac{x^3}6+o\left(x^3\right)}{x^3}=$
$\quad=\lim\limits_{x\to0}\left[-\dfrac16+\dfrac{o\left(x^3\right)}{x^3}\right]=-\dfrac16\;.$
Moreover , by letting $\;y=\dfrac{\sin x}x-1\;,\;$ we get that
$\left(\dfrac{\sin x}x\right)^{\frac1{x^2}}=\bigg[\big(1+y\big)^{\frac1y}\bigg]^{\frac y{x^2}}=\bigg[\big(1+y\big)^{\frac1y}\bigg]^{\frac{\sin x-x}{x^3}}.\quad\color{blue}{(*)}$
Since $\;y=\dfrac{\sin x}x-1\;,\;$ it results that
$\lim\limits_{x\to0}\;y=\lim\limits_{x\to0}\left(\dfrac{\sin x}x-1\right)=1-1=0\;.$
From $\;(*)\;,\;$ it follows that
$\lim\limits_{x\to0}\left(\dfrac{\sin x}x\right)^{\frac1{x^2}}=\lim\limits_{x\to0}\bigg[\big(1+y\big)^{\frac1y}\bigg]^{\frac{\sin x-x}{x^3}}=$
$\quad=\bigg[\lim\limits_{x\to0}\big(1+y\big)^{\frac1y}\bigg]^{\lim\limits_{x\to0}\frac{\sin x-x}{x^3}}=$
$\quad=\bigg[\lim\limits_{y\to0}\big(1+y\big)^{\frac1y}\bigg]^{\lim\limits_{x\to0}\frac{\sin x-x}{x^3}}\!=e^{-\frac16}\;,$
hence ,
$\lim\limits_{x\to0}\left(\dfrac{\sin x}x\right)^{\frac1{x^2}}=e^{-\frac16}\;.$
