I haven't proved continuity in ages, but I found a situation where it would be helpful.
I'd like to prove the Riemann-Liouville differintegral https://en.wikipedia.org/wiki/Riemann%E2%80%93Liouville_integral is a continuous operator in its integration parameter $\alpha$ because I'm pretty sure that it is, but I haven't seen a proof for it and it seems like a good exercise anyway.
The way I'd argue this is the following:
Let $f(x)$ be continuous over $[a,b]$ and denote $I^{\alpha}f$ as the $\alpha$th order differintegral as defined in wikipedia. ( Do we restrict $x$ to being nonnegative?) and $\alpha \geq 0$ and $\beta \geq 0.$
Then for every $\epsilon >0,$ there exists $\delta > 0$ such that for all nonnegative $\alpha$, $0 < |\alpha - \beta| < \delta$ implies $|I^{\alpha}f - I^{\beta}f| < \epsilon.$
Now how do we find an $\epsilon$ in terms of $|\alpha - \beta|$ or something comparable to that expression?
Well, from the definition we have
$$|I^{\alpha}f - I^{\beta}f| = | \frac{1}{\Gamma(\alpha)} \int_{a}^{x}f(t)(x-t)^{\alpha} - \frac{1}{\Gamma(\beta)} \int_{a}^{x}f(t)(x-t)^{\beta}|$$
By assumption, $f$ is continuous so it attains its minimum and maximum, and we may then denote $M = \sup( |f| )$ to obtain
$$| \frac{1}{\Gamma(\alpha)} \int_{a}^{x}f(t)(x-t)^{\alpha} - \frac{1}{\Gamma(\beta)} \int_{a}^{x}f(t)(x-t)^{\beta}| \leq |\frac{1}{\Gamma(\alpha)} \int_{a}^{x}|M(x-t)^{\alpha}| - \frac{1}{\Gamma(\beta)} \int_{a}^{x}|M(x-t)^{\beta}||$$
Then at this point I'm getting confused and stuck because the $|x-a|^{\alpha}$ term does not return a nice expression when you integrate it, unlike $(x-a)^{\alpha}.$