Best way find $\lim_{x\to 0}( \frac {\sin x}{x})^{\frac 1x}$ $\lim_{x\to 0}( \frac {\sin(x)}{x})^{\frac 1x}$ $$$$
I can use Tailor to get to $\lim_{x\to 0}(1+\epsilon(x))^\frac 1x$ $$$$  $(\epsilon(x)\underset{x\to\infty}\to 0)   $
$$$$
but does that mean that the limit equals 1?
 A: What you did is not enough, since the exponent goes to infinity as the stuff in parenthesis goes to one. We'll have to be smarter! Here's how to approach problems like this, with variables in the exponent.
Use the fact that $\ln(x)$ is continuous. Since $\ln()$ is continuous, that means we can interchange $\ln()$ and $\lim()$ in any problem:
$$
\ln(\lim[f(x)]) = \lim(\ln[f(x)])
$$
In the particular case of this problem, we can say
$$
\lim(\frac{\sin(x)}{x})^{1/x}=\exp\{\ln[\lim(\frac{\sin(x)}{x})^{1/x}]\} = \exp\{\lim[\ln(\frac{\sin(x)}{x})^{1/x}]\}
$$
Then you just need to figure out
$$
L=\lim_{x\rightarrow 0}(\ln[(\frac{\sin(x)}{x})^{1/x}])
$$
which you can do just by standard manipulations of $\ln$ and l'Hopital's rule. Then take $\exp\{L\}$, and that's your answer!
Note: $\exp\{x\}=e^{x}$, I just wrote it as $\exp$ to make it easier to read.
A: In this answer, it is shown that
$$
\cos(x)\le\frac{\sin(x)}{x}\le1
$$
Therefore,
$$
\cos(x)^{1/x}\le\left(\frac{\sin(x)}x\right)^{1/x}\le1
$$
Since
$$
\begin{align}
\lim_{x\to0}\cos(x)^{1/x^2}
&=\lim_{x\to0}\left(1-2\sin^2(x/2)\right)^{1/x^2}\\
&=\lim_{x\to0}\left(1\color{#C00000}{-\frac{\sin^2(x/2)}{2(x/2)^2}}x^2\right)^{1/x^2}\\[3pt]
&=e^{\color{#C00000}{-1/2}}
\end{align}
$$
we have
$$
\begin{align}
\lim_{x\to0}\cos(x)^{1/x}
&=\lim_{x\to0}e^{-x/2}\\
&=1
\end{align}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{x\to0}\left(\frac{\sin(x)}x\right)^{1/x}=1
$$
A: "Best way" is slightly vague. If you need an answer which can be written in small amount of time (say for a competitive exam) then I guess you can write
\begin{align}
\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/x} &= \lim_{x \to 0}\exp\left\{\frac{1}{x}\log\left(\frac{\sin x}{x}\right)\right\}\notag\\
&= \lim_{x \to 0}\exp\left\{\frac{1}{x}\log\left(\frac{1}{x}\left(x - \frac{x^{3}}{3!} + o(x^{3})\right)\right)\right\}\notag\\
&= \lim_{x \to 0}\exp\left\{\frac{1}{x}\log\left(1 - \frac{x^{2}}{6} + o(x^{2})\right)\right\}\notag\\
&= \lim_{x \to 0}\exp\left\{\frac{1}{x}\left(- \frac{x^{2}}{6} + o(x^{2})\right)\right\}\notag\\
&= \lim_{x \to 0}\exp\left(-\frac{x}{6} + o(x)\right)\notag\\
&= \exp(0) = 1\notag
\end{align}
Note that the symbols $o(x), o(x^{2}),\cdots$ should always be used when applying Taylor series expansions for evaluation of limits.
However if you need an answer which focuses more on concepts then you can prooceed as follows. Let $L$ be the desired limit and then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{\sin x}{x}\right)^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left\{1 + \left(\frac{\sin x}{x} - 1\right)\right\}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\left(\frac{\sin x}{x} - 1\right)\dfrac{\log\left\{1 + \left(\dfrac{\sin x}{x} - 1\right)\right\}}{\dfrac{\sin x}{x} - 1}\notag\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}}\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = \frac{\sin x}{x} - 1)\notag\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{2}}\cdot 1\notag\\
&= 0\notag
\end{align}
Hence $L = e^{0} = 1$. The evaluation of $$\lim_{x \to 0}\frac{\sin x - x}{x^{2}}$$ is given here.
A: Ok I got it:
made a mistake with the Taylor remainder 
Sin(x) translates to (x + $x\epsilon(x))$ so $$$$
$\lim_{x\to 0}(1+x\epsilon(x))^\frac 1x= \lim_{x\to 0}(1+\frac 1{\frac 1{x\epsilon(x)}})^\frac 1x$=...= $e^0=1$
A: \begin{align}
\lim_{x\to 0}( \frac {\sin(x)}{x})^{\frac 1x}&=\lim_{x\to 0}( \prod_{n=1}^{\infty}\Big(1-\frac{x^2}{n^2\pi^2}\Big))^{\frac 1x}\\
&=\prod_{n=1}^{\infty}\lim_{x\to 0} \Big(1-\frac{x^2}{n^2\pi^2}\Big)^{\frac 1x}\\
&=\prod_{n=1}^{\infty}1=1
\end{align}
A: If you are able to find
$$
L=\lim_{x\to0}\log\biggl(\left(\frac{\sin x}{x}\right)^{\!1/x}\,\biggr)=
\lim_{x\to0}\frac{\log\frac{\sin x}{x}}{x},
$$
where $f(x)=\frac{\sin x}{x}$, then you're done, because your original limit will be $e^L$ (or $0$ if $L=-\infty$ or $\infty$ if $L=\infty$).
Now the limit is just the derivative at $0$ of the function $\log f(x)$, where
$$
f(x)=\begin{cases}
\dfrac{\sin x}{x} & \text{if $x\ne0$}\\[6px]
\;1 & \text{if $x=0$}
\end{cases}
$$
that admits the Taylor series representation
$$
f(x)=1-\frac{x^2}{3!}+\frac{x^4}{4!}-\dots+
(-1)^n\frac{x^{2n}}{(2n+1)!}+\dotsb
$$
that we can deduce from the Taylor series for $\sin x$. Therefore, using again the Taylor development,
$$
\lim_{x\to0}\frac{\log f(x)}{x}=\lim_{x\to 0}\frac{-x^2/6+o(x^2)}{x}=0
$$
and your original limit is $e^0=1$.
