Limit of $\lim\limits_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$ How to evaluate the limit: $$\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$$
I think it goes to $1$ because $\lim\limits_{x \to 0} \frac{\sin{x}}{x} =1$ and so power of $1$ should also be $1$. Am I right?
 A: You are wrong. For example, $$\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}=e$$
But by your approach it would be $1$. This is because $1^\infty$ is undefined and not $1$. The correct approach would be to use the limit I gave as example to get:
$$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^3}}=\lim_{x\to 0}\left(1+\frac{\sin x-x}{x}\right)^{\frac{x}{\sin x-x}\frac{1}{x^3}\frac{\sin x-x}{x}}=\lim_{x\to 0}\,\,\,e\,\,^{\frac{\sin x-x}{x^4}}$$
You can continue from here.
A: Edit. Turns out this limit does not exist, because the one-sided limits as $x \to 0+$ and $x \to 0-$ are different. See @Américo's answer for more details.
The problem in your idea will become apparent once you take logs: Define $L$ to be the limit. Then
$$
\ln L = \lim_{x \to 0} \frac{\ln(\frac{\sin x}{x})}{x^3} .
$$
Now, it is true that the numerator approaches $0$ as $x \to 0$. But I cannot immediately conclude that the limit is $0$, since this is a $0/0$ indeterminate form. Can you spot the connection to your question here? 
I presume you should be more comfortable in evaluating $0/0$ indeterminate forms. Can you take it from here? As a starting point, you might want to get rid of the $\ln$ by using some standard limit theorems. (Hint: $\ln(1+h)$ as $h$ goes to $0$.) 
A: Hint: Rewrite $\left( \dfrac{\sin x}{x}\right)
^{1/x^{3}}$ as  $$\left( \frac{\sin x}{x}\right)
^{1/x^{3}}=e^{\left( \ln \left( \frac{\sin x}{x}
\right) \right) /x^{3}}$$
and look at the side limits: $$\lim_{x\rightarrow 0^{+}}e^{\left( \ln \left( \frac{\sin x}{x}\right)
\right) /x^{3}}\neq \lim_{x\rightarrow 0^{-}}e^{\left( \ln \left( \frac{
\sin x}{x}\right) \right) /x^{3}}.$$
Additional hint: show that
$$\lim_{x\rightarrow 0^{+}}\left( \ln \left( \frac{\sin x}{x}\right) \right)
/x^{3}=-\infty ,$$
$$\lim_{x\rightarrow 0^{-}}\left( \ln \left( \frac{\sin x}{x}\right) \right)
/x^{3}=+\infty .$$
A: Hint: If allowed, I would use the Maclaurin series for $\sin(x)=x+\frac{x^3}{6}+O(x^5)$. Then, after diving by $x$, note that $\lim\limits_{x\to0}(1+x^2)^{1/x^3}=\lim\limits_{x\to0}(1+x^2)^{(1/x^2)(1/x)}=\lim\limits_{x\to0}e^{1/x}$.
