Calculate exponential limit involving trigonometric functions Calculate the following limit:
$$\lim_{x \rightarrow 0} \left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}$$
I know the result must be $\sqrt[3]{e}$ but I don't know how to get it. I've tried rewriting the limit as follows:
$$\lim_{x \rightarrow 0} e ^ {\ln {\left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}}} = \lim_{x \rightarrow 0} e ^ {\frac{1}{\sin^2 x} \ln {\left( \frac{\tan x}{x} \right)}}$$
From this point, I applied l'Hospital's rule but got $1$ instead of $\sqrt[3]{e}$.
Thank you!

You have an indeterminate form of kind $1^{\infty}.$ You can solve this type of problems via:
$$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} (f(x)-1)g(x)}.$$
Edit
If $\lim_{x\to a}f(x)=1$ then it is $$\lim_{x\to a} f(x)^{g(x)}=\lim_{x\to a} (1+f(x)-1)^{\dfrac{f(x)-1}{f(x-1)}g(x)}=\lim_{x\to a} \left( (1+f(x)-1)^{\dfrac{1}{f(x)-1}}\right)^{f(x-1)g(x)}.$$
Now, it is $$\lim_{x\to a} \left( (1+f(x)-1)^{\dfrac{1}{f(x)-1}}\right)=e.$$
So, if $\lim_{x\to a} (f(x)-1)g(x)$ exists we have $$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} (f(x)-1)g(x)}.$$
End of the edit
In this case,
$$\begin{align}\lim_{x\to 0}\left(\dfrac{\tan x}{x}-1\right)\dfrac{1}{\sin^2 x} & \\ &= \lim_{x\to 0}\dfrac{\tan x-x}{x\sin^2 x} \\ &=\lim_{x\to 0}\dfrac{\dfrac{1}{\cos^2 x}-1}{\sin^2 x+2x\sin x\cos x}
\\& =\lim_{x\to 0}\dfrac{\sin^2 x}{\cos^2 x(\sin^2 x+2x\sin x\cos x)}\\&=\lim_{x\to 0}\dfrac{\sin x}{\cos^2 x(\sin x+2x\cos x) }\\&
=\lim_{x\to 0}\dfrac{\cos x}{-2\cos x\sin x(\sin x+2x\cos x)+\cos^2x(3\cos x-2x\sin x) }\\&=\dfrac 13.\end{align}$$
