Limit Brownian Bridge Integral As a solution of the Brownian Bridge SDE, we arrive at the solution
\begin{align}
X_t = (1-t) \int_0^t \frac{1}{1-s}\ dB_S
\end{align}
defined for $0 \leq t <1$. In order to show that for any $g \in C[0,1]$ 
\begin{align}
\lim_{t \uparrow 1}\ (1-t) \int_0^t \frac{g(s)}{(1-s)^2}\ ds = g(1),
\end{align}
I am considering two cases.
Case 1:
\begin{align} 
\lim_{t \uparrow 1}\ \int_0^t \frac{g(s)}{(1-s)^2}\ ds = \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty,
\end{align}
such that we apply l'Hôpital's rule and find that
\begin{align}
\lim_{t \uparrow 1}\ \frac{\frac{g(t)}{(1-t)^2}}{\frac{1}{(1-t)^2}} = g(1).
\end{align}
Case 2:
\begin{align} 
\lim_{t \uparrow 1}\ \int_0^t \frac{g(s)}{(1-s)^2}\ ds \neq \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty.
\end{align}
Now, clearly
\begin{align}
(*) = \lim_{t \uparrow 1}\ \frac{ \int_0^t \frac{g(s)}{(1-s)^2}\ ds }{\frac{1}{1-t}} \to 0 \qquad \text{since} \qquad \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty.
\end{align}
However, how to show that $(*) \to g(1)$ as well?
 A: 
NOTE: 
Referring to the note at the end of the General Proof Section of THIS ARTICLE, L'Hospital's Rule states that given two functions $f$ and $g$ that are differentiable in an open neighborhood of $\xi$ with $g'(x)\ne 0$ in that neighborhood, such that $\lim_{x\to \xi}|g(x)|=\infty$, then if the limit $\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$ exists, whether finite or infinite, then we have
$$\lim_{x\to \xi}\frac{f(x)}{g(x)}=\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$$
Note carefully that nowhere in the statement of L'Hospital's Rule is the assumption that $\lim_{x\to \xi}f(x)$ even exists.


Now, I thought it might be instructive to present an approach that does not use L'Hospital's Rule.  Rather, we use the fact that $g(s)$ is continuous on $[0,1]$, and therefore uniformly continuous there.  
As a first step, we write
$$\begin{align}
(1-t)\int_0^t \frac{g(s)}{(1-s)^2}\,ds&=(1-t)\left(\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds+g(t)\int_0^t \frac{1}{(1-s)^2}\,ds\right)\\\\
&=t\,g(t)+(1-t)\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\tag 1
\end{align}$$
Second, from the uniform continuity of $g$, given any $\epsilon>0$, there exists a number $\delta>0$ such that $|g(s)-g(t)|<\epsilon$ whenever $t-\delta <s<t$.  For that $\delta$, we now take $1> t>1-\delta$.  Then, we have
$$\begin{align}
\left|(1-t)\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\right|&=\left|(1-t)\int_0^{t-\delta} \frac{g(s)-g(t)}{(1-s)^2}\,ds+(1-t)\int_{t-\delta}^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\right|\\\\
&\le (1-t)\int_0^{t-\delta} \frac{|g(s)-g(t)|}{(1-s)^2}\,ds+(1-t) \int_{t-\delta}^t \frac{|g(s)-g(t)|}{(1-s)^2}\,ds\\\\
&\le \frac{2||g||_{\infty}(1-t)(1-\delta)}{\delta}+\epsilon\left(1-\frac{(1-t)}{\delta}\right) \tag 2
\end{align}$$
Third, letting $t\to 1^-$ in $(2)$ we see that for any $\epsilon>0$, $\lim_{t\to 1^-}\left|(1-t)\int_0^t \frac{g(s)-g(1)}{(1-s)^2}\,ds\right|<\epsilon$, and hence the limit is zero.  
Using this result in $(1)$ results in the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{t\to 1^-}(1-t)\int_0^t \frac{g(s)}{(1-s)^2}\,ds=g(1)}$$
And we are done! 
