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 $e$.
 A: Your solution is good until the final step.
$$\lim_{x\to \infty}{x\over x+1}=\lim_{x\to \infty}{x\over x(1+{1\over x})}=1$$
A: Your original answer of the unedited first post is good. Note that $L = \displaystyle \lim_{x \to \infty} \dfrac{1}{\left(1+\frac{1}{x}\right)^{x}} =....=\dfrac{1}{e}$. Since you edited it, we got a new question. In this case, write
$\left(1+\dfrac{1}{x+1}\right)^x = \dfrac{\left(1+\dfrac{1}{1+x}\right)^{1+x}}{1+\dfrac{1}{1+x}}\to \dfrac{e}{1} = e$ as $x \to \infty$.
A: Another way to do it : consider $$A={\Big(1+\frac {1}{x+1}}\Big)^x$$ Take logarithms $$\log(A)=x\log\Big(1+\frac {1}{x+1}\Big)$$ Now, using the fact that, for small values of $y$, $\log(1+y)=y+O\left(y^2\right)$, replace $y$ by $\frac {1}{x+1}$ and get $$\log(A)\sim\frac {x}{x+1}$$ Then $\log(A)\to 1$ and $A\to e$.
Using the same method as in the textbook $$A=\frac{{\Big(1+\frac {1}{x+1}}\Big)^{x+1}}{1+\frac 1{x+1}}$$ Change $x=y-1$; so the numerator is familiar to you and the denominator tends to $1$.
A: Notice, let $x+1=t\implies t\to \infty$ as $x\to \infty$ $$\lim_{x\to \infty}\left(1+\frac{1}{x+1}\right)^x$$
$$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t-1}$$
$$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\cdot \left(1+\frac{1}{t}\right)^{-1}$$
$$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\cdot \lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{-1}$$
 $$=(e)\cdot (1)^{-1}$$
$$=\color{red}{e}$$
A: 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 $\lim_{x \rightarrow \infty}\frac {x}{x + 1} = {1 \over e}$
