Limit of $\frac{(x-2)n^n}{(n+1)^n(n+1)}$ for $n$ tends to infinity I know the limit  $$\lim_{n \to \infty}\frac{(x-2)n^n}{(n+1)^n(n+1)} = 0$$  but I can't seem to properly get to that answer. Could anyone help me out?
 A: Note that since $\frac{n}{n+1}<1$, we have for any $\epsilon>0$
$$\begin{align}
\left|\frac{(x-2)n^n}{(n+1)^n\,(n+1)}\right|&=\left|\frac{(x-2)}{\,(n+1)}\,\underbrace{\left(\frac{n}{n+1}\right)^n}_{<1}\right|\\\\
&< \frac{|x-2|}{n+1}\\\\
&<\frac{|x-2|}{n}\\\\
&<\epsilon
\end{align}$$
whenever $n>\frac{|x-2|}{\epsilon}$.
A: Hint:
$$\frac{n^n}{(n+1)^n}=\left(1-\frac1{n+1}\right)^n\to \frac1e$$
A: $$ \frac{(x-2)n^n}{(n+1)^n(n+1)} =\frac{(x-2)}{(n+1)} \left(\frac{n}{n+1}\right)^n=\frac{(x-2)}{(n+1)} \frac{1}{\left(1+\frac1n\right)^n}\to0\cdot\frac1e=0$$
A: Let $\theta$ be $n+1$. Then, if $n\rightarrow\infty$, so does $\theta$. This observation is going to be useful later in the solution.
\begin{align}
\lim_{n\to\infty}\frac{(x-2)n^n}{(n+1)^n(n+1)}
&=\lim_{n\to\infty}\frac{(x-2)n^n}{(n+1)^n(n+1)}\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n^n}{(n+1)^n}\cdot\frac{1}{n+1}\right)\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n^n}{(n+1)^{n+1}}\cdot\frac{n}{n}\right)\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n^{n+1}}{(n+1)^{n+1}}\cdot\frac{1}{n}\right)\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n+1}\cdot\lim_{n\to\infty}\frac{1}{n}\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right)^{n+1}\cdot 0\\
&=(x-2)\cdot\lim_{n\to\infty}\left(\frac{n+1}{n+1}-\frac{1}{n+1}\right)^{n+1}\cdot 0\\
&=(x-2)\cdot\lim_{n\to\infty}\left(1+\frac{-1}{n+1}\right)^{n+1}\cdot 0\\
&=(x-2)\cdot\lim_{\theta\to\infty}\left(1+\frac{-1}{\theta}\right)^{\theta}\cdot 0\\
&=(x-2)\cdot e^{-1} \cdot 0\\
&= 0
\end{align}
