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. 
 A: Using the fact that, for $x \to 0$,
$$\begin{align}
1 - \cos x &= \frac{x^2}2 + o(x^2)\\
\tan x &= x + \frac{x^3}3 + o(x^3)\\
\ln(1 + x) &= x + o(x),
\end{align}$$
we can rewrite the limit as
$$\begin{align}\lim_{x \to 0}\exp\left(\frac1{1 - \cos x}\ln\left(\frac{\tan x}x\right)\right) &= \lim_{x \to 0}\exp\left(\frac2{x^2}\ln\left(1 + \frac{x^2}3\right)\right) =\\
&= \exp\left(\frac23\right) = \sqrt[3]{e^2}.
\end{align}$$
A: Using the limits
$$
\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=\lim_{x\to0}\frac{\log(1+x)}{x}=1
$$
and one application of L'Hospital, we get
$$
\begin{align}
&\lim_{x\to0}\frac1{1-\cos(x)}\log\left(\frac{\tan(x)}x\right)\\
&=\lim_{x\to0}\frac{1+\cos(x)}{\sin^2(x)}\log\left(1+\frac{\tan(x)-x}x\right)\\
&=\lim_{x\to0}\frac{1+\cos(x)}{\sin^2(x)}\frac{\tan(x)-x}x\frac{\log\left(1+\frac{\tan(x)-x}x\right)}{\frac{\tan(x)-x}x}\\
&=\lim_{x\to0}\underbrace{\frac{(1+\cos(x))x^2}{\sin^2(x)}}_2\frac{\tan(x)-x}{x^3}\underbrace{\frac{\log\left(1+\frac{\tan(x)-x}x\right)}{\frac{\tan(x)-x}x}}_1\\
&=2\lim_{x\to0}\frac{\tan(x)-x}{x^3}\\[9pt]
&=2\lim_{x\to0}\frac{\tan^2(x)}{3x^2}\\[9pt]
&=\frac23
\end{align}
$$
Therefore,
$$
\lim_{x\to0}\left(\frac{\tan(x)}x\right)^{\Large\frac1{1-\cos(x)}}=e^{2/3}
$$
A: Using Series Expansion,
$\displaystyle\tan x=x+\frac{x^3}3+O(x^5),\cos x=1-\dfrac{x^2}2+O(x^4),$
$$\left(\frac{\tan x}x\right)^{\dfrac1{1-\cos x}}=\left(1+\frac{x^2}3+O(x^4)\right)^{\dfrac1{1-\cos x}}$$
$$=\displaystyle\left[\left(1+\frac{x^2}3+O(x^4)\right)^{\frac1{\frac{x^2}3+O(x^4)}}\right]^\frac{\dfrac{x^2}3+O(x^4)}{\dfrac{x^2}2+O(x^4)}$$
Remember $\lim_{h\to0}(1+h)^{1/h}=e$
A: 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
$$
\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{1-\cos x}} = \lim_{x\to 0} \exp\left[\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x}\right] = e^{2/3}.
$$
A: use the facts $$\tan x = x + \frac13x^3 + \cdots, \cos x = 1 - \frac12 x^2 + \cdots , (1 + small)^{BIG} = e^{small \times BIG} + \cdots $$  so that $$ \left(\frac{\tan x}{x} \right)^{\frac{1}{1-\cos x}} = \left(1 + \frac13 x^2 + \cdots\right)^{\frac2{x^2} + \cdots} = e^{\frac{2}{3}} + \cdots \to e^{2/3} \text{ as }x \to 0$$
