2015/082) Evaluate $(\dfrac{n}{n+1})^{1 + n}$ as $n\rightarrow \infty$

$(\dfrac{n}{n+1})^{1 + n}$

Let's remember one thing first: lim $x\rightarrow\infty (1 + \frac{r}{x})^x = e^r$
$(\dfrac{n}{n+1})^{1 + n}$

=  $(1- \dfrac{1}{n+1})^{1 + n}$

substitute x = n + 1

Now, as n goes to infinity, then n + 1 will also go to infinity, and x will go to infinity

lim $n\rightarrow\infty (1 - \dfrac{1}{n + 1})^{n + 1}$ 

= lim $x\rightarrow\infty (1 - \dfrac{1}{x})^x$ 

 
Which is $e^{-1}$, or $\dfrac{1}{e}$