Find the limit of $(\tan(x) + \sec (x))^{1/\sin(x)}$ $$ \lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = e $$
how to arrive at e, according to wolfram alpha, that this is the answer?

Note that:
$$\begin{align}\lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = &\lim_{x \rightarrow 0} \left[\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right]^{1/\sin x} =\\ &\lim_{x \rightarrow 0} \frac{[1+\sin x]^{1/\sin x}}{[\cos x]^{1/\sin x}} =\\
&\frac{\lim_{x \rightarrow 0}[1+\sin x]^{1/\sin x}}{\lim_{x \rightarrow 0}[\cos x]^{1/\sin x}} =\\
&\frac{e}{1}=e.\end{align}$$
