Limit Function: Finding Solution

[tmt1]

I'm trying to find the limit of this function:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}}$$

I can simplify it to this:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n}(n + 1)}$$

But I'm not sure of the best way to proceed.

 

[Prove It]

I'm trying to find the limit of this function:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}}$$

I can simplify it to this:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n}(n + 1)}$$

But I'm not sure of the best way to proceed.

$\displaystyle \begin{align*} &= \lim_{n \to \infty} \left( \frac{n}{n + 1} \right) ^n \left( \frac{1}{n + 1} \right) \\ &= \lim_{n \to \infty} \left( \frac{n}{n + 1} \right) ^n \, \lim_{n \to \infty} \frac{1}{n + 1} \\ &= \lim_{n \to \infty} \left( 1 - \frac{1}{n + 1} \right) ^n \,\lim_{n \to \infty} \frac{1}{n + 1} \\ &= \lim_{n \to \infty} \mathrm{e}^{ \ln{ \left[ \left( 1 - \frac{1}{n + 1} \right) ^n \right] } }\,\lim_{n \to \infty} \frac{1}{n + 1} \\ &= \mathrm{e}^{ \lim_{n \to \infty} \frac{\ln{ \left( 1 - \frac{1}{n + 1} \right) }}{\frac{1}{n}} }\,\lim_{n \to \infty} \frac{1}{n + 1} \end{align*}$

Use L'Hospital's Rule to finish it off :)

 

[Evgeny.Makarov]

An alternative to L'Hospital's rule in this context is the fact that $\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$. Then
\begin{align}
\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n&=
\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^{n+1}\left(\frac{n+1}{n}\right)\\
&=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)\\
&=e^{-1}\cdot1.
\end{align}

 

[I like Serena]

Alternatively:
$$\frac{n^n}{({n + 1})^{n + 1}} = \left(\frac{n}{n + 1}\right)^{n} \cdot \frac{1}{n + 1} \le 1^n \cdot \frac{1}{n + 1} \to 0
$$
So:
$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}} = 0$$