Proving that a moment generating function converges pointwise I have found a moment generating function $M_n$ given by
$\cfrac{(1-e^t)e^{\frac tn}}{n(1-e^{\frac tn})}$ if $t\ne 0$
and 1 if $t =0$
How do I prove that $M_n$ converges point-wise to the moment generating function of a random variable that is uniformly distributed on $(0,1)$?
 A: The moment generating function of a uniform random variable on $[a,b]$ is given by
\begin{align*}
M(t)=\begin{cases}\frac{e^{tb}-e^{ta}}{t(b-a)}&\text{if }t\neq0;\text{ and}\\
1&\text{if }t=0.\end{cases}
\end{align*}
Thus,
your problem amounts to showing that
$$\lim_{n\to\infty}\frac{(1-e^t)e^{t/n}}{n(1-e^{t/n})}=\frac{e^t-1}{t}$$
whenever $t$ is a fixed real number other than $0$.
We can break this problem down in several sub-problems:

First, we find $\lim_{n\to\infty}(1-e^t)e^{t/n}$ for some $t\neq 0$.
Let $t\neq 0$ be fixed.
Clearly,
$t/n$ converges to zero as $n\to\infty$.
Thus,
since the function $x\mapsto e^x$ is continuous,
it follows that $e^{(t/n)}\to e^0=1$ as $n\to\infty$.
Since $(1-e^t)$ is constant,
we conclude:
$$\lim_{n\to\infty}(1-e^t)e^{t/n}=1-e^t.$$

Second,
we find $\lim_{n\to\infty}n(1-e^{t/n})$.
If we try this directly,
we obtain $\infty\cdot 0$,
which is an indeterminate form.
To solve this,
we can use a clever trick:
$$\lim_{n\to\infty}n(1-e^{t/n})=\lim_{n\to\infty}\frac{1-e^{t/n}}{1/n},$$
which gives $0/0$.
This is still an indeterminate form,
but you can apply l'Hôpital's rule to get rid of the indetermination.
This should give you that 
$$\lim_{n\to\infty}\frac{1-e^{t/n}}{1/n}=\lim_{n\to\infty}-te^{t/n}=-t.$$

Finally,
use the fact that,
if two limits $\lim_n x_n$ and $\lim_n y_n$ exist and are finite,
then
$$\lim_{n\to\infty}\frac{x_n}{y_n}=\frac{\lim_{n\to\infty}x_n}{\lim_{n\to\infty}y_n}.$$
