What is $\lim\limits_{n\to\infty}\frac{x}{n(1-(1-x/n)e^{t/n})}$ I'm interested in finding this limit, and while I can plug it into Wolfram and have it spit out $\frac{x}{x-t}$ for me, I'm interested in how to "do" this limit.
I'd post my efforts, but I honestly haven't really been able to get farther than
\begin{align*}
\displaystyle\lim_{n\to\infty}\dfrac{x}{n\left(1-\left(1-\dfrac{x}{n}\right)e^{t/n}\right)}=\lim_{n\to\infty}\frac{x}{n-ne^{t/n}+xe^{t/n}}
\end{align*}
If anyone could point me in the right direction, that'd be cool. I'm just curious!

If you are OK to use the limit 
\begin{equation*}
\lim_{y\rightarrow 0}\frac{e^{y}-1}{y}=1
\end{equation*}
as a basic limit (which can be proved by LHR) then you can write the
original fraction as
\begin{eqnarray*}
\frac{x}{n(1-(1-\frac{x}{n})e^{t/n})} &=&\frac{x}{t}\frac{\frac{t}{n}}{(1-(1-%
\frac{x}{t}\frac{t}{n})e^{t/n})} \\
&=&\frac{x}{t}\frac{\frac{t}{n}}{e^{t/n}(e^{-t/n}-(1-\frac{x}{t}\frac{t}{n}))%
} \\
&=&\frac{x}{t}\frac{e^{-t/n}}{\frac{(e^{-t/n}-1+\frac{x}{t}\frac{t}{n}))}{%
\frac{t}{n}}} \\
&=&\frac{x}{t}\frac{e^{-t/n}}{\frac{(e^{-t/n}-1)}{\frac{t}{n}}+\frac{x}{t}}
\end{eqnarray*}
then passing to the limit one gets
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{x}{n(1-(1-\frac{x}{n})e^{t/n})}\overset{%
y=t/n}{=}\lim_{-y\rightarrow 0}\frac{x}{t}\frac{e^{-y}}{\frac{(e^{-y}-1)}{%
-(-y)}+\frac{x}{t}}=\frac{x}{t}\frac{e^{0}}{\left( -1+\frac{x}{t}\right) }=%
\frac{x}{x-t}.\ \blacksquare 
\end{equation*}
