<s>[INST] 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. 
 [/INST] [\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} ]
</s>