<s>[INST] Limit of trig function raised to trig function 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.
 [/INST] 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 $$\frac{6480-7 \pi ^2}{6480 \sqrt[3]{e}}\approx 0.708892$$
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