Evaluating $\lim_\limits{x \to 0 }(\frac{\tan x}{x})^{\frac{1}{x^2}}$ Any ideas on how to tackle this limit?
$$\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$$
I tried many ways but only got more complex stages, not easier ones...
 A: Notice that
$$\tan x=x+\frac{x^3}3+o(x^3)$$
so
$$\left(\frac{\tan x}{x}\right)^{1/{x^2}}\sim_0\left[\left(1+\frac{x^2}{3}\right)^{3/x^2}\right]^{1/3}\xrightarrow{x\to0}e^{1/3}\tag{$\because [1+1/x]^x\sim_0e$}$$
A: $$\begin{align}\lim_{x \to0 }\left(\frac{\tan x}{x}\right)^{1/x^2}
&=\lim_{x \to0 }e^{\displaystyle \left(\frac1{x^2}\ln\frac{\tan x}x\right)}
\\&=\lim_{x \to0 }e^{\displaystyle \left(\frac1{x^2}\frac{\ln\left(1+\left(\frac{\tan x}x-1\right)\right)}{\left(\frac{\tan x}x-1\right)}\left(\frac{\tan x}x-1\right)\right)}
\\&=\lim_{x\to0}e^{\displaystyle \left(\frac1{x^2}\left(\frac{\tan x}x-1\right)\right)}\tag{$\because \lim_{x\to0}\frac{\ln(1+ x)}x=1$}
\\&=\lim_{x\to0}e^{\displaystyle \left(\frac{\tan x-x}{x^3}\right)}\tag{$*$, proved below}
\\&=e^{1/3}
\end{align}$$
Now to prove $(*)$ we can use L'Hospital:
$$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\sec^2x-1}{3x^2}=\lim_{x\to0}\frac{2\sec^2 x\tan x}{6x}=\lim_{x\to0}\frac13\underbrace{\frac{\tan x}{x}}_1\underbrace{\sec^2x}_1=\frac13$$
Or Taylor:
$$\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{(x+x^3/3+O(x^5))-x}{x^3}=\frac13$$
A: By far the easiest and fastest way:
$$\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = \lim_{x \to 0}\exp\left(\frac1{x^2}\ln\left(1 + \frac{x^2}{3}\right)\right) = \lim_{x \to 0}\exp\left(\frac1{x^2}\cdot\frac{x^2}3\right) = e^{1/3}$$
Using the fact that, for $x \to 0$,
$$\begin{align}
\tan x &\sim x + \frac{x^3}3\\
\ln(1 + x) &\sim x
\end{align}$$
A: First, subtracting $(10)$ from $(9)$ in this answer, we get that
$$
\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13\tag{1}
$$
Next, since $\lim\limits_{x\to0}\left(1+tx^2\right)^{1/x^2}=e^t$ converges uniformly on compact sets,
$$
\begin{align}
\lim_{x\to0}\left(\frac{\tan(x)}x\right)^{1/x^2}
&=\lim_{x\to0}\left(1+\frac{\tan(x)-x}{x^3}x^2\right)^{1/x^2}\\
&=\lim_{x\to0}\left(1+\frac13x^2\right)^{1/x^2}\\[6pt]
&=e^{1/3}\tag{2}
\end{align}
$$
