Determining the limit of $\bigl(1 + \frac{1}{x+1}\bigr)^x$ as $x\to\infty$ $$L = \lim_{x\to \infty} \biggl(1+\frac{1}{x+1}\biggr)^x $$

This one has to solved using $$\lim_{x\to \infty} \biggl(1+\frac{1}{x}\biggr)^x = e.$$   

I did this
\begin{align}L &amp;= \lim_{x\to \infty}\biggl( \biggl(1+\frac{1}{x+1}\biggr)^{x+1}\biggr)^{\frac{x}{x+1}}\\
&amp; = e^{\frac{x}{x+1}}
\end{align}

I get the limit to be (I figured it out this is wrong) $$\frac{1}{e}$$ 
while the answer is $e$.
Your solution is good until the final step.
$$\lim_{x\to \infty}{x\over x+1}=\lim_{x\to \infty}{x\over x(1+{1\over x})}=1$$
Well that new question is entirely different.

remember if $\lim b_n = B$ and $\lim c_n = C \ne 0$ then $\lim b_n/c_n = B/C$

$\lim (1 + 1/x)^x = e$

so $\lim (1 + {1 \over {x + 1}})^{x+1} = e$

and $\lim (1 + {1 \over {x + 1}}) = 1$

so $\lim (1 + {1 \over {x + 1}})^x = $

$\lim \frac { (1 + {1 \over {x + 1}})^{x+1}}{1 + {1 \over {x + 1}}} =$

$\frac {\lim (1 + {1 \over {x + 1}})^{x+1}}{\lim (1 + {1 \over {x + 1}})}  = {e \over 1} = e$

==== original answer for the original (miswritten) question below ====

Well, I thought my way was simple.  

For $1 + 1/x = \frac {x + 1}{x}$

So $\lim_{x \rightarrow \infty}\frac {x + 1}{x} = e \ne 0$

So $\lim_{x \rightarrow \infty}\frac {x}{x + 1} = {1 \over e}$