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?

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 $1$.
