Find the limit of $(\tan(x) + \sec (x))^{1/\sin(x)}$

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 $$(\tan(x)+\sec(x))^{\csc(x)} = \exp(1+O(x)) = e + O(x)$$

Let $x=\arcsin(t)$. Then, we have

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

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Now, letting $u=1/t$ reveals

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

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Finally, enforcing the substitution $v=(u-1)/2$ yields

$$\left(\tan(x)+\sec(x)\right)^{1/\sin(x)}=\left(1+\frac1{v}\right)^{v}\sqrt{1+\frac1{v}}$$

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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!