Limit of a simple Integral Need to show that $$\lim_{t\to 1^-}(1-t)\int_0^t\frac{g(s)}{(1-s)^2}ds=g(1)$$ for any continuous function $g(s)$.
I tried a variable change $u=1/(1-s)$ which gives $du=ds/(1-s)^2$ with $$(1-t)\int_1^{\frac{1}{1-t}}g(1-\frac{1}{u})du$$
I cannot seem to figure out how to get to the limit, given that thus far it's correct.

According to Theorem 5.13 in Rudin's "Principles of mathematical analysis" (slightly restated)

If $f$ and $h$ are real and differentiable in $(0,1)$ and if $h'(x)\neq 0$ in a neighborhood of $1$, and $h(x)\to +\infty$ as $x\to 1^-$, and in addition $f'(x)/h'(x)\to A$ as $x\to 1^-$, then $f(x)/h(x)\to A$ as $x\to 1^-$.

Applying this to
$$
f(x)=\int_0^x\frac{g(s)}{(1-s)^2}\,ds\quad\text{and}\quad h(x)=\frac{1}{1-x},
$$
we get (here we use the fundamental theorem of calculus and the property that $g$ is continuous)
$$
\frac{f'(x)}{h'(x)}=\frac{g(x)/(1-x)^2}{1/(1-x)^2}=g(x)\to g(1)\quad (x\to 1^-).
$$
Hence, the limit you look for (check the conditions) is also $g(1)$.
