<s>[INST] Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Is there a way to do it without differentiating so many times? That is exhausting and confusing and will probably cause errors. I would really appreciate your help with this. 
 [/INST] Using basic Taylor series as $x \to 0$,
$$
\frac{1}{1-\cos x} = \frac{1}{1-(1-\frac{x^2}{2} + o(x^2))} \sim \frac{2}{x^2}
$$
and
$$
\log\left(\frac{\tan x}{x}\right) = \log\left(1 + \frac{x^2}{3} + o(x^2)\right) \sim \frac{x^2}{3}.
$$
Therefore
$$
\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x} \sim \frac{2}{x^2}\frac{x^2}{3} = \frac{2}{3}
$$
and the result follows by taking exponentials
$$
\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{1-\cos x}} = \lim_{x\to 0} \exp\left[\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x}\right] = e^{2/3}.
$$
</s>