# 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!

• Please don't use TeX-only titles and refrain from using displaystyle in titles. The former causes technical issues with some users and the latter makes it take up too much vertical space on the front page. – AlexR May 20 '15 at 7:46
• @AlexR Apologies. Noted for the future. – Old mate May 20 '15 at 8:23

Hint

Just consider $$A=\frac{x}{n\left(1-\left(1-\frac{x}{n}\right)e^{t/n}\right)}$$ and now use the fact that, for small $y$ $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ Replace $y$ by $\frac tn$ and simplify.

I am sure that you can take from here.

• Let me research big O notation (I haven't come across it yet in my studies and am not entirely sure how to deal with it), and I'm sure I will be able to have a crack at it. Thank you! :) – Old mate May 20 '15 at 6:17
• You are very welcome. Since you did learnt it yet, forget big O notation and just use the fact that, when $y$ is small, $e^y\approx 1+y$. This is enough for the problem. – Claude Leibovici May 20 '15 at 6:24
• Ah, I see. Thank you very much, to both you and @Olivier Oloa. – Old mate May 20 '15 at 6:25

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*}

Hint. You may write, as $n \to \infty$, \begin{align} e^{t/n}&=1+\frac tn+\mathcal{O}\left(\frac{t^2}{n^2}\right) \end{align}

giving \begin{align} 1-\left(1-\dfrac{x}{n}\right)e^{t/n}&=1-\left(1-\dfrac{x}{n}\right)\left(1+\frac tn++\mathcal{O}\left(\frac{t^2}{n^2}\right)\right)\\\\ &=\dfrac{x-t}{n}+\mathcal{O}\left(\frac{t^2}{n^2}\right)\\\\ \end{align} \begin{align} n\left(1-\left(1-\dfrac{x}{n}\right)e^{t/n}\right)&=x-t+\mathcal{O}\left(\frac{t^2}{n}\right) \end{align} thus $$\frac{x}{n\left(1-\left(1-\frac{x}{n}\right)e^{t/n}\right)}=\frac{x}{x-t}+\mathcal{O}\left(\frac{t^2}{n}\right) \to \frac{x}{x-t}.$$