Evaluating a limit expression I am trying to show the  following identity but I am stuck.
$$\lim_{t\nearrow 1}(1-t)\int_0^t\frac{g(s)}{(1-s)^2}\,ds = g(1)$$
for any $g \in C[0,1]$. Apparently, the proof follows from L'Hopital's rule but I guess for that I first need to show 
$$\lim_{t\nearrow 1}\left\lvert\int_0^t\frac{g(s)}{(1-s)^2}\,ds \right\rvert= \infty$$
But that is not true since I can just take $g(s) = (1-s)^2$. But then the main result itself still holds. Is the hint about L'Hopital's rule wrong then? If so, how do I prove this result? I would appreciate any help. Thanks in advance.
 A: Let $G(t):=\int_0^t\frac{g(s)}{(1-s)^2}\,ds$. Note that $\lim_{t\nearrow 1}\frac{1}{1-t}=\infty$. Thus, by the first fundamental theorem of calculus and L'Hospital's rule, we have (for $t\in (0,1)$)
$$\lim_{t\nearrow 1}(1-t)G(t) = \lim_{t\nearrow 1}\frac{G(t)}{1/(1-t)}=\lim_{t\nearrow 1}\frac{G^{'}(t)}{1/(1-t)^2}=\lim_{t\nearrow 1}\frac{g(t)/(1-t)^2}{1/(1-t)^2}=g(1)$$
A: You may use Weierstrass approximation theorem, providing a polynomial $p(s)$ for which:
$$\forall s\in[0,1],\quad \left|g(s)-p(s)\right|\leq \varepsilon,\quad p'(0)=0, \tag{1}$$
then, for every $t\in[0,1)$,
$$(1-t)\left|\int_{0}^{t}\frac{g(s)}{(1-s)^2}\,ds-\int_{0}^{t}\frac{p(s)}{(1-s)^2}\,ds\right|\leq\varepsilon \tag{2}$$
and now we are allowed to use integration by parts:
$$ \int_{0}^{t}\frac{p(s)}{(1-s)^2}\,ds = \left.\frac{p(s)}{(1-s)}\right|_{0}^{t}-\int_{0}^{t}\frac{p'(s)}{(1-s)}\,ds\tag{3}$$
then multiply both sides by $(1-t)$ and take the limit as $t\to 1^-$ to prove that the original limit is arbitrarily close to $g(1)$. Notice that, since $p'(s)=s\cdot q(s)$:
$$\left|\int_{0}^{t}\frac{p'(s)}{(1-s)}\,ds\right|^2 = \left|\int_{0}^{t}\frac{s\cdot q(s)}{(1-s)}\,ds\right|^2\leq \int_{0}^{t}\frac{s\,ds}{(1-s)^2}\int_{0}^{t}s\cdot q(s)^2\,ds\tag{4}$$
hence:
$$\left|\int_{0}^{t}\frac{p'(s)}{(1-s)}\,ds\right|\leq\frac{C_q}{\sqrt{1-t}}.\tag{5}$$
