Let $f(x)=(\sin^2 x)^{\sec^2 x}$. Where is this function defined? Can it be defined at the missing points in order to make it continuous there?
Edit: Is there a way to show this using the fact that $x^a = e^{alnx}$? I'd like to know. My answer below does not use this fact.
Firstly, note that $f(x)$ is undefined whenever $sec^2 x$ is undefined or $sin^2 x=sec^2 x=0$. $sin^2 x=sec^2 x=0\Rightarrow sin^2 xcos^2x=1=0$, which is obviously false. Thus, $f(x)$ is undefined exactly when $sec^2 x$ is undefined. $sec^2 x$ is undefined exactly when $x=\frac{\pi}{2}+n\pi,n\in \mathbb{Z}$. Since $cos^2 x=0$ at all of these points, to make $f(x)$ continuous, we only need to find $lim_{x\to \frac{\pi}{2}} (sin^2 x)^{sec^2 x}$ and set $f(x)$ equal to it to make $f(x)$ continuous at each point where it originally was discontinuous. Set $y=\frac{1}{cos^2 x}$. Then $y\to\infty \Leftrightarrow x\to (\frac{\pi}{2}+n\pi,n\in\mathbb{Z})$, so $lim_{x\to \frac{\pi}{2}} (sin^2 x)^{sec^2 x}=lim_{x\to\frac{\pi}{2}} (1-cos^2 x)^{\dfrac{1}{cos^2 x}}=lim_{y\to\infty} (1-\dfrac{1}{y})^y=e^{-1}$.
Thus, we can define $f(x)$ as a piecewise function where $f(x)=e^{-1}$ exactly when $x=\dfrac{\pi}{2}+n\pi,n\in\mathbb{Z}$ and $f(x)=(sin^2 x)^{sec^2 x}$ otherwise.