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 &= \lim_{x\to \infty}\biggl( \biggl(1+\frac{1}{x+1}\biggr)^{x+1}\biggr)^{\frac{x}{x+1}}\\
& = 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.
$$\lim_{x\to \infty}{x\over x+1}=\lim_{x\to \infty}{x\over x(1+{1\over x})}=1$$
