Here's a little question I saw in a book recently, which I can see but can't set out my own formal proof and it's annoying me.
Say $\lim_{x\to\infty} g(x) = a$
and $g$ is continuous,
so now prove that
$\lim_{x\to\infty} \frac{1}{x} \int_0^x g(y) \mathrm{d}y = a$ .
I can see that if we define $\int_0^x g(y)\mathrm{d}y = G(x) - G(0)$
then $\frac{1}{x} \int_0^x g(y) \mathrm{d}y = \frac{G(x) - G(0)}{x}$
which clearly looks a lot like the limit of a differential, but I'm not sure how to handle the limit?!