limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of 
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws. 
 A: $$\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\frac x{x+2}\cdot\left(1-\frac1{x+2}\right)^{x+2}\left(1-\frac1{x+2}\right)^{-1}\xrightarrow[x\to\infty]{}1\cdot\frac1e\cdot1=\frac1e$$
A: $$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\dfrac{(x+1)^{x+2}-(x+1)^{x+1}}{(x+2)^{x+2}}$$
$$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\Big(1-\dfrac{1}{x+2}\Big)^{x+2}\Big(\dfrac{x+2}{x+1}\Big)-\Big(1-\dfrac{1}{x+2}\Big)^{x+1}\dfrac{1}{x+2}\to e^{-1}+0$$
A: [\begin{gathered}
  \mathop {\lim }\limits_{x \to \infty } \frac{{x{{\left( {x + 1} \right)}^{x + 1}}}}
{{{{\left( {x + 2} \right)}^{x + 2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{x}
{{x + 2}}.\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 1}}
{{x + 2}}} \right)^{x + 1}} = 1.\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1}
{{x + 2}}} \right)^{x + 1}} \hfill \\
   = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{1}
{{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{ - \frac{{x + 1}}
{{x + 2}}}} = {\left[ {\mathop {\lim }\limits_{x \to \infty } {{\left( {1 - \frac{1}
{{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{\mathop {\lim }\limits_{x \to \infty }  - \frac{{x + 1}}
{{x + 2}}}} = {e^{ - 1}} = \frac{1}
{e} \hfill \\ 
\end{gathered} ]
A: Start with$$\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\frac{x}{x+2}\cdot\left(\frac{x+1}{x+2}\right)^{x+1},$$then note that$$\left(\frac{x+2}{x+1}\right)^{x+1}=\left(1+\frac{1}{x+1}\right)^{x+1}\longrightarrow e.$$
A: $$\lim_{x\rightarrow \infty }\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\\\lim_{x\rightarrow \infty }\frac{(x+1)^{x+1}}{(x+2)^{x+1}}*\frac{x}{x+2}=\\\lim_{x\rightarrow \infty }(\frac{x+1}{x+2})^{x+1}\frac{x}{x+2}=\\(1-\frac{1}{x+2})^{x+1}* \lim_{x\rightarrow \infty }\frac{x}{x+2}=\\e^{-1}*\lim_{x\rightarrow \infty }\frac{x}{x+2}\\e^{-1}*1\\=\frac{1}{e}$$
A: let $y = \dfrac{x(x+1)^{x+1}}{(x+2)^{x+2}},$  then 
$\begin{align} \ln y &= \ln x + (x+1)\ln(x+1)-(x+2)\ln(x+2)\\
 &=\ln x +(x+1)[\ln x + \ln(1 + 1/x)] -(x+2)[\ln x+ \ln(1 + 2/x)]\\
&=\left(1+x+1-x-2  \right)\ln x +(x+1)\left(1/x  + \cdots\right)-(x+2)\left(2/x +\cdots\right)\\
&=-1 + \cdots
\end{align}$
so $$\lim_{x \to \infty} \ln y= -1 \text{ and } \lim_{x \to \infty} y= \frac{1}{e}.$$
