How to show that $\lim_{n \rightarrow \infty} (\frac{n}{n+a})^n = \frac{1}{e^a}$? How to show that $$\lim_{n \rightarrow \infty} (\frac{n}{n+a})^n = \frac{1}{e^a}\,\,?$$
If $a=1\,$ then $$(\frac{n}{n+1})^n = (\frac{1}{\frac{n+1}{n}})^n = \frac{1}{(\frac{n+1}{n})^n} \rightarrow \frac{1}{e}$$ as $ n \rightarrow \infty$
 A: Hint: Write your function as $\bigg(\frac{n}{n+a}\bigg)^n=\bigg(\frac{1}{1+\frac{a}{n}}\bigg)^n=\frac{1}{(1+\frac{a}{n})^n}$
A: Rewrite your function as $$\displaystyle e ^{-n\ln \left(1+\cfrac an\right)}$$ and do a change of variable $t=\cfrac 1n$ where $t\rightarrow 0$ when $n\rightarrow +\infty$ then apply l'Hopital's.
A: Note that - 
$$\frac{n}{n+a}=\frac{n+a-a}{n+a}=1-\frac{a}{n+a}$$
And therefore for any $a\in\mathbb{R}$:
$$\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^n=\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^{n+a}\cdot\left(1-\frac{a}{n+a}\right)^{-a}=$$$$=\lim_{n\to\infty} \left(1-\frac{a}{n+a}\right)^{n+a}\cdot\lim_{n\to\infty}\left(1-\frac{a}{n+a}\right)^{-a}=e^{-a}$$
A: for this case we can calculate Ln() of the function and then result is exp() of the answer.
$$  \lim_{n \rightarrow \infty} (\frac{n}{n+a})^n  $$
so we have :
$$ \text{Ans}= \lim_{n \rightarrow \infty}\left[\ln \left(\frac{n}{n+a}\right)^n \right] =\lim_{n \rightarrow \infty} \left[ n. \ln \left(\frac{n}{n+a}\right) \right] = \infty . 0   $$
$$ \text{Ans}= \lim_{n \rightarrow \infty}  \left[ \frac{\ln(\frac{n}{n+a})}{\frac{1}{n}} \right]=\frac{0}{0}  \space \rightarrow \space \text{Ambiguity in Mathematics}$$
we can use Hopital's rule,so we have:
$$  \text{Ans}=\lim_{n \rightarrow \infty}\left[  \frac{\frac{a}{n(n+a)}}{\frac{-1}{n^2}} \right]=\lim_{n \rightarrow \infty} \left[ \frac{-an^2}{n(n+a)} \right]=\lim_{n \rightarrow \infty} \left[ \frac{-an}{(n+a)} \right] = -a $$
$$  \lim_{n \rightarrow \infty} (\frac{n}{n+a})^n  = e^{\text{Ans}} =e^{-a}=\frac{1}{e^a}$$
