Find $\lim_{x \to 0} (\frac{\tan(x)}{x})^{\frac{1}{x}}$ 
Find $$\lim_{x \to 0} (\frac{\tan(x)}{x})^{\frac{1}{x}}$$

My first idea to solve this was to try to evaluate it and then apply the L'Hospital's rule. This is what I managed to achieve:
$$\left(\frac{\tan(x)}{x}\right)^\frac{1}{x}=e^{\frac{1}{x}\ln\left(\frac {\tan(x)}x \right)} $$
However, this is problematic, because the L'Hospital's rule is not applicable in the exponent, therefore my transformation was not a good shot. What transformation should I apply to get the desired form?
 A: $$\tan x=x+O(x^3)$$
$$\frac{\tan x}x=1+O(x^2)$$
$$\ln\frac{\tan x}x=O(x^2)$$
$$\ln\left[\left(\frac{\tan x}x\right)^{1/x}\right]=\frac1x\ln\frac{\tan x}x=O(x)$$
etc.
A: \begin{align*}
\lim_{x\rightarrow 0}\dfrac{1}{x}\ln\left(\dfrac{\tan x}{x}\right)&=\lim_{x\rightarrow 0}\dfrac{1}{x}\ln\left(\dfrac{x+\dfrac{1}{3}x^{3}+\cdots}{x}\right)\\
&=\lim_{x\rightarrow 0}\dfrac{1}{x}\ln\left(1+\dfrac{1}{3}x^{2}+\cdots\right)\\
&=\lim_{x\rightarrow 0}\dfrac{1}{x}\left(\dfrac{1}{3}x^{2}-\dfrac{1}{2}\left(\dfrac{1}{3}x^{3}\right)^{2}+\cdots\right)\\
&=0.
\end{align*}
So the limit is $e^{0}=1$.
A: Easy trick
$$\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac1{x}}  =\lim_{x\to 0}\exp\left(\frac{1}{x}\ln\left(\frac{\tan x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{x}{3}\frac{\ln\left(1+\frac{x^2}{3}\right)}{\frac{x^2}{3}}\right)= \color{blue}{1}$$
Given that $$\tan x -x \sim \frac{x^3}{3}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1+h\right)}{h} = 1$$
A: i would write $$\exp\left({\frac{\ln(\tan(x))-\ln(x)}{x}}\right)$$
A: By Taylor's series:
$\tan x=x+\frac{x^3}{3}+o(x^3)$
$\log (1+x)=x+o(x)$
Thus:
$$\left(\frac{\tan x}{x}\right)^\frac{1}{x}=e^{\frac{1}{x}\ln\left(\frac {\tan x }x \right)}=e^{\frac{1}{x}\ln\left(1+\frac{x^2}{3}+o(x^2)\right)}=e^{\frac{x}{3}+o(x)}\to e^0=1$$
A: If you want to use L'Hospital's rule, I would note that the limit of these two functions are equal:
$\frac{ln(\frac{tanx}{x})}{x}$ and $\frac{(tan^2x+1)x-tanx}{xtanx}$, and then rewrite the second term as $tanx + \frac{x-tanx}{xtanx}$, noting that $tanx$ is eliminated. It remains to find $\frac{x-tanx}{xtanx}$. But by applying L'Hospital's rule again, we can see that this is equivalent to $\frac{-tanx}{1+\frac{x}{tanx}(1+tan^2x)}$, which goes to $0$ since $\frac{x}{tanx}$ goes to $1$. So your limit is $e^0$ which is $1$.
A: Let us prove the result by using only l'Hospital's rule. We have
$${\left(\frac{\tan  \left(x\right)}{x}\right)}^{\frac{1}{x}} = \exp  \left(\frac{\ln \left|\tan  \left(x\right)\right|-\ln  \left|x\right|}{x}\right) := \exp  \left(\frac{f \left(x\right)}{x}\right)$$
Note that $f \left(0\right) = 0$ and
$${f'} \left(x\right) = \frac{1}{\tan  \left(x\right)} \left(1+{\tan }^{2} \left(x\right)\right)-\frac{1}{x} = \tan  \left(x\right)-x \left(\frac{x}{\tan  \left(x\right)}\right) \left(\frac{\tan  x-x}{{x}^{3}}\right)$$
By l'Hospital rule we have
$$
 \frac{\tan  \left(x\right)}{{x}} \mathop{\longrightarrow}\limits_{x \rightarrow  0} 1,
\qquad  \frac{\tan  \left(x\right)-x}{{x}^{3}} \mathop{\longrightarrow}\limits_{x \rightarrow  0} \frac{1}{3}$$
it follows that ${f'} \left(x\right) \mathop{\longrightarrow}\limits_{x \rightarrow  0} 0$, hence applying again
l'Hospital's rule we get 
$$\frac{f \left(x\right)}{x} \mathop{\longrightarrow}\limits_{x \rightarrow  0} 0 \quad  \Longrightarrow  \quad  {\left(\frac{\tan  \left(x\right)}{x}\right)}^{\frac{1}{x}} \mathop{\longrightarrow}\limits_{x \rightarrow  0} 1$$
