Calculate the limit of $\lim _{n \to \infty} (1 - \frac{1}{n+1})^n $ I need to calculate the limit of
$$\lim _{n \to \infty} (1 - \frac{1}{n+1})^n $$
I know that it should be $e^{-1} $ because of 
$\lim _{n \to \infty} (1 + \frac{k}{n})^n = e^k $ but how can I show that the $+1$ downstairs doesn’t change that. Intuitively speaking it's clear, but how can I show it?
 A: Notice that $$\left(1-\dfrac{1}{n+1}\right)^n=\left(1-\dfrac{1}{n+1}\right)^{n+1}\left(1-\dfrac{1}{n+1}\right)^{-1}$$ and now apply the product rule for limits. We have $$\lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)^{n+1}=e^{-1}$$ and $$\lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)^{-1}=1$$ so $$\lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)^n=\lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)^{n+1}\lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)^{-1}=e^{-1}$$
A: HINT: it is $$\left(\frac{n+1-1}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
A: Hint:
$$\left(1 - \frac{1}{n+1}\right)^n = \left(1 - \frac{1}{n+1}\right)^{n+1} \left(1 - \frac{1}{n+1}\right)^{-1}$$
A: For any finite $k$, we have
$$\lim _{n \to \infty} \left(1 - \frac{1}{n+k}\right)^{n}$$
$$ = \lim _{n \to \infty} \frac{\left(1 - \frac{1}{n+k}\right)^{n+k}}{\left(1 - \frac{1}{n+k}\right)^k}$$
$$ = \frac{\lim _{n \to \infty} \left(1 - \frac{1}{n+k}\right)^{n+k}}{\lim _{n \to \infty} \left(1 - \frac{1}{n+k}\right)^k}$$
$$ = \frac{e^{-1}}{1}$$
$$ = \boxed{\frac{1}{e}}$$
The limit of the denominator is obviously $1$ in the fraction above. For a rigorous proof use Bernoulli's inequality.
A: We can prove that as follows:
$$\begin{align*}\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n=&\lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right)^n=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\\
=&\lim_{n\to\infty}\left(\frac{1}{\frac{n+1}{n}}\right)^n=\lim_{n\to\infty}\frac{1}{\left(\frac{n+1}{n}\right)^n}=\lim_{n\to\infty}\frac{1}{\left(1+\frac{1}{n}\right)^n}\overset{*}{=}\\
=&\frac{1}{\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}
\end{align*}$$
*This equality holds, since the limit $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ does exist and is equal to $e$.
Generally, when you don't know how to compute a limit, brute-force may work! ;)
A: Write
$$
(1 - \frac{1}{n+1})^n
=\frac{(1 - \frac{1}{n+1})^{n+1}}{(1 - \frac{1}{n+1})}\to {1/e}
$$
since the denominator tends to $1$.
