limit of $f(x) = \lim \limits_{x \to 0} (\frac{\sin x}{x})^{\frac 1x}$ Any ideas how to calculate this limit without using taylor?
 $$f(x) = \lim \limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac1x}$$
 A: Take the log of both sides and examining the 2 sides of the limits yields
$$\ln L^+=\lim \limits_{x \to 0^+} \frac{\ln (\sin x)-\ln(x)}{x}$$
$$\ln L^-=\lim \limits_{x \to 0^-} \frac{\ln (\sin (-x))-\ln(-x)}{x}$$
which can be solved by L'Hopital's to both equal $0$, so the limit is 1.
A: With Taylor expansions: (just for reference)
We will use 
$$\begin{align}
\sin u &= u + o(u^2) \\
\ln(1+x) &= u  + o(u)
\end{align}$$
when $u\to0$. (In particular, $\ln(1+o(u)) = o(u)$.)

Write
$$
\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}
= \left(\frac{x+o(x^2)}{x}\right)^{\frac{1}{x}}
= \left(1+o(x)\right)^{\frac{1}{x}}
= e^{ \left(\frac{1}{x}\ln\left(1+o(x)\right)\right) }
= e^{ \left(\frac{1}{x}\left(o(x)\right)\right) }
= e^{o(1) }
$$
so the limit is $e^0=1$.
A: Notice, $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp \left(\frac{1}{x}\ln\left(\frac{\sin x}{x}\right)\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{\ln(\sin x)-\ln(x)}{x}\right)$$
using L'Hosptal's rule for $\frac 00$ form, 
$$=\lim_{x\to 0}\exp \left(\frac{\frac{\cos x}{\sin x}-\frac1x}{1}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{x\cos x-\sin x}{x\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{x(-\sin x)+\cos x-\cos x}{x\cos x+\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\sin x}{x\cos x+\sin x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\cos x-\sin x}{-x\sin x+\cos x+\cos x}\right)$$
$$=\lim_{x\to 0}\exp \left(\frac{-x\cos x-\sin x}{-x\sin x+2\cos x}\right)$$
$$=\exp \left(\frac{0}{0+2}\right)=e^0=\color{red}{1}$$
A: HINT.-$$ (\frac{\sin x}{x})^{1/x}=\left([1+(\frac{\sin x}{x}-1)]^{\frac{1}{\frac{\sin x}{x}-1)}}\right)^{\frac{\sin x-x}{x^2}}$$
It follows $$\lim \limits_{x \to 0} (\frac{\sin x}{x})^{1/x}=e^{\lim {x\to 0}\frac{\sin x -x}{x^2}}=e^0=1$$
A: Here's a proof
that just uses
basic properties
of
$\sin, \cos$,
and
$\ln$.
Since
$-1 \le \cos(x) < 1$
and $\sin'(x) = \cos(x)$
and $\cos'(x) = -\sin(x)$,
$\sin(x)
=-\int_0^x \cos(t) dt
$
so
$|\sin(x)|
\le |x|
$.
Also,
since,
for $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
=\int_0^{x} \frac{dt}{1+t}
$,
$\ln(1+x)
\le x
$
and,
for $1 > x > 0$,
$\ln(1-x)
=\int_1^{1-x} \frac{dt}{t}
=-\int^1_{1-x} \frac{dt}{t}
=-\int^0_{-x} \frac{dt}{1+t}
$,
so,
if $\frac12 > x > 0$,
$-\ln(1-x)
=\int^0_{-x} \frac{dt}{1+t}
\ge \frac{x}{1-x}
\ge 2x
$.
Therefore,
for
$-\frac12 < x < 1$,
$|\ln(1+x)|
\le 2|x|
$.
$\begin{array}\\
\cos(x)
&=1-2\sin^2(x/2)\\
&=1+O(x^2)
\qquad\text{with the implied constant
being less than 1},\\
\text{so that}\\
\frac{\sin x}{x}
&=\frac1{x}\int_0^x \cos(t)dt\\
&=\frac1{x}\int_0^x (1+O(t^2))dt\\
&=\frac1{x} (t+O(t^3))|_0^x\\
&=\frac1{x} (x+O(x^3))\\
&=1+O(x^2))\\
\text{so}\\
\ln(\frac{\sin x}{x})
&=\ln(1+O(x^2))\\
&=O(x^2)\\
\text{so}\\
\frac1{x}\ln(\frac{\sin x}{x})
&=\frac1{x}(O(x^2))\\
&=O(x)\\
&\to 0 \text{ as }x \to 0\\
\text{so}\\
(\frac{\sin x}{x})^{1/x}
&\to 1\\
\end{array}
$
