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?
 A: As $ x \to 0$, $\tan(x) + \sec(x) = 1 + x + O(x^2)$ while $\csc(x) = 1/x + O(x)$.
Thus $$\ln\left((\tan(x)+\sec(x))^{\csc(x)}\right) = \csc(x) \ln(\tan(x)+\sec(x))
= \left(\frac{1}{x} + O(x)\right)(x + O(x^2)) = 1 + O(x)$$
and $$(\tan(x)+\sec(x))^{\csc(x)} = \exp(1+O(x)) = e + O(x)$$
A: Let $x=\arcsin(t)$.  Then, we have
$$\begin{align}
\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}&=\left(\frac{1+t}{\sqrt{1-t^2}}\right)^{1/t}\\\\
&=\left(\frac{1+t}{1-t}\right)^{1/(2t)}\\\\
\end{align}$$

Now, letting $u=1/t$ reveals
$$\begin{align}
\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}&=\left(\frac{u+1}{u-1}\right)^{u/2}\\\\
&=\left(1+\frac{2}{u-1}\right)^{u/2}
\end{align}$$

Finally, enforcing the substitution $v=(u-1)/2$ yields
$$\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}=\left(1+\frac1{v}\right)^{v}\sqrt{1+\frac1{v}}$$

As $x\to 0$, $v\to \infty$ and we have
$$\lim_{x\to 0}\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}=\lim_{v\to\infty}\left(1+\frac1{v}\right)^{v}\sqrt{1+\frac1{v}}=e$$
as was to be shown!
A: Note that:
$$\begin{align}\lim_{x \rightarrow 0} [\tan(x) + \sec(x)]^{\csc(x)} = &\lim_{x \rightarrow 0} \left[\frac{\sin x}{\cos x} + \frac{1}{\cos x}\right]^{1/\sin x} =\\ &\lim_{x \rightarrow 0} \frac{[1+\sin x]^{1/\sin x}}{[\cos x]^{1/\sin x}} =\\
&\frac{\lim_{x \rightarrow 0}[1+\sin x]^{1/\sin x}}{\lim_{x \rightarrow 0}[\cos x]^{1/\sin x}} =\\
&\frac{e}{1}=e.\end{align}$$
A: $$(\sec x+\tan x)^{\csc x}=\left(\left(1+\sec x+\tan x-1\right)^{1/(\sec x+\tan x-1)}\right)^{(\sec x+\tan x-1)/\sin x}$$
Use https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution for the outer exponent
A: Using the same approach as in Robert Israel's anwser, consider
$$A=(\tan (x)+\sec (x))^{\csc (x)}\implies \log(A)=\csc (x)\log (\tan (x)+\sec (x))$$ Now, using Taylor series built at $x=0$
$$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$
$$\sec(x)=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^5\right)$$
$$\tan (x)+\sec (x))=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{5 x^4}{24}+O\left(x^5\right)$$
$$\log (\tan (x)+\sec (x))=x+\frac{x^3}{6}+O\left(x^5\right)$$
$$\csc(x)=\frac{1}{x}+\frac{x}{6}+\frac{7 x^3}{360}+O\left(x^5\right)$$
$$\log(A)=1+\frac{x^2}{3}+O\left(x^4\right)$$
$$A=e^{\log(A)}=e+\frac{e x^2}{3}+O\left(x^4\right)$$ which is a quite good approximation of the original function (just for suriosity, plot both of them on the same graph for $0 \leq x \leq 0.5$).
Try fro $x=\frac \pi {6}$ for which we know the exact values of the trigonometric functions. The exact value is
$A=3$ while the approximation gives $e+\frac{e \pi ^2}{108}\approx 2.9667$.
