I need help on evaluating this limit:
$\lim_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{1-\cos{x}}}$. So far I have tried setting the limit to L and taking the ln of both sides. So $\ln{L}=\lim_{x\to0}(\frac{1}{1-\cos{x}}\ln{(\frac{\sin{x}}{x})})$. I then tried using L'Hopital on the indeterminate 0/0 to no luck. Any hints would be appreciated.
$$L=\lim_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{1-\cos{x}}}$$
$$\ln L=\lim_{x\to0}\frac{\ln (\frac{\sin{x}}{x})}{1-\cos{x}}$$
$$\ln L=\lim_{x\to0}\frac{\frac{d}{dx}\ln (\frac{\sin{x}}{x})}{\frac{d}{dx}(1-\cos{x})}$$
$$\ln L=\lim_{x\to0}\frac{\frac{d}{dx}(\ln (\sin x)-\ln x)}{\sin{x}}$$
$$\ln L=\lim_{x\to0}\frac{\frac{1}{\tan x}-\frac{1}{x}}{\sin{x}}$$
$$\ln L=\lim_{x\to0}\frac{x-\tan x}{x\tan x\sin{x}}$$
If you are allowed to use series expansion, consider $$A=\left(\frac{\sin{(x)}}{x}\right)^{\frac{1}{1-\cos{(x)}}}\implies \log(A)=\frac{1}{1-\cos{(x)}}\log\left(\frac{\sin{(x)}}{x}\right)$$ and use $$\sin{(x)}=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ $$\frac{\sin{(x)}}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$ $$\log\left(\frac{\sin{(x)}}{x}\right)=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(A)=\frac{-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right) }{\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right) }=\frac {-\frac{1}{6}-\frac{x^2}{180}+O\left(x^4\right) }{\frac{1}{2}-\frac{x^2}{24}+O\left(x^4\right)}$$ which already shows the limit for $\log(A)$.
If you want to go further, use the long division (or Taylor series again) to get $$\log(A)=-\frac{1}{3}-\frac{7 x^2}{180}+O\left(x^4\right)$$ and $$A=e^{\log(A)}=e^{-\frac{1}{3}}\left(1-\frac{7 x^2}{180} \right)+O\left(x^4\right)$$ which shows the limit and how it is approached.
It is even a good approximation of the function. If you use $x=\frac \pi 6$ the exact value would be $$A=\left(\frac{3}{\pi }\right)^{2 \left(2+\sqrt{3}\right)}\approx 0.708768$$ while the truncated series would give $$\frac{6480-7 \pi ^2}{6480 \sqrt[3]{e}}\approx 0.708892$$
Hint:
$$((1+\frac{sin x-x}x)^{\frac x{\sin x-x}})^{\frac{sin x-x}{x(1-\cos x)}}$$
Now use Are all limits solvable without L'Hôpital Rule or Series Expansion for the exponent.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that $\ds{\,\mrm{sinc}:\mathbb{R} \to \mathbb{R}}$ is an even function such that $\ds{\,\mrm{sinc}'}$ is an odd function, $\ds{\,\mrm{sinc}''}$ is an $\ds{\color{#f00}{even}}$ function, $\ds{\,\mrm{sinc}'''}$ is an odd function and so on.
$\ds{\,\mrm{sinc}\pars{x} = \left\{\begin{array}{lcl} \ds{\sin\pars{x} \over x} & \mbox{if} & \ds{x \not = 0} \\ \ds{1} & \mbox{if} & \ds{x = 0} \end{array}\right.}$
\begin{align} \lim_{x \to 0}\bracks{\sin\pars{x} \over x}^{1/\bracks{1-\cos\pars{x}}} & = \exp\pars{\lim_{x \to 0}{\ln\pars{\mrm{sinc}\pars{x}} \over 1 - \cos\pars{x}}} = \exp\pars{\lim_{x \to 0}{{\mrm{sinc}'\pars{x}} \over \sin\pars{x}\,\mrm{sinc}\pars{x}}} \\[5mm] & = \exp\pars{\lim_{x \to 0}{{\mrm{sinc}'\pars{x}} \over \sin\pars{x}}} = \exp\pars{\lim_{x \to 0}{{\mrm{sinc}''\pars{x}} \over \cos\pars{x}}} \\[5mm] & = \exp\pars{\lim_{x \to 0}{\mrm{sinc}''\pars{x}}} = \exp\pars{\lim_{x \to 0}\bracks{-\,{1 \over 3} + {x^{2} \over 10} + \,\mrm{O}\pars{x^{4}}}} \\[5mm] & = \bbx{\expo{-1/3}} \approx 0.7165 \end{align}
