Proving the limit of $\left(\frac n{n+1}\right)^n$is $e^{-1}$

Question

Find the limit of $\lim_{n \to \infty}({n\over n+1})^n$.

$$\lim_{n \to \infty}({n\over n+1})^n=\lim_{n \to \infty}({n+1-1\over n+1})^n=\lim_{n \to \infty}(1-{1\over n+1})^n$$

What Should I do next?

Answer 1

$$ ({n\over n+1})^n={1 \over ({n+1\over n})^n} = {1 \over (1+{1\over n})^n} \to {1 \over e} $$

Answer 2

$$\begin{align}\lim_{n \to \infty}({n\over n+1})^n=\lim_{n \to \infty}({n+1-1\over n+1})^n&=\lim_{n \to \infty}(1-{1\over n+1})^n\\&=\underbrace{\lim_{n\to\infty}\left[\left(1-\frac{1}{n+1}\right)^{-(n+1)}\right]^{-\frac{n}{n+1}}}_{e^{-\frac{n}{n+1}}}\\&=\frac{1}{e}\end{align}$$

Answer 3

Observing that the exponential and logarithm are inverses of each other, you can apply L'Hopital's rule to get: \begin{equation*} \lim_{n\to\infty}(\frac{n}{n+1})^n=\exp \lim_{n\to\infty}(n\ln(\frac{n}{n+1})) =\exp (\lim_{n\to\infty}\frac{\ln(\frac{n}{n+1})}{\frac{1}{n}}) \\ = \exp(\lim_{n\to\infty} -\frac{n}{n+1}) =\exp(\lim_{n\to\infty}\frac{-1}{1+\frac{1}{n}})=\frac{1}{e}. \end{equation*}