How to find $\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}$ I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?
$$\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x[(x)(1+\frac{1}{x})]^x}{[(x)(1+\frac{1}{x})]^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}(1+\frac{1}{x})^x}{x^x(1+\frac{1}{x})^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}e}{x^xe} $$
$$\lim_{x \to \infty} \frac{x-x}{1} $$
$$\lim_{x \to \infty} \frac{0}{1} $$
$$0$$
I understand my mistakes may be simple and trivial, but I'm trying to learn. Thank you for your help!

This solution is not as elegant as @farruhotas, but it's more direct. This solution relies on Taylor approximations and handwaving. I realize it's not rigorous; the purpose of this is providing intuition as to why the limit is $\frac{1}{2}$. (I'll be assuming that $x$ is an integer. Assuming the limit exists, since $x \to \infty$, this isn't an issue.)
Consider the numerator. The highest power terms are $x^{x+1}$ terms. We will first show that the $x^{x+1}$ terms cancel as $x \to \infty$. Consider the expansion of $(x+1)^x$:
$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$ 
Its highest power terms are of order $x^x$. Since the $x^k$ term of $\binom{x}{k}$ is approximately $\frac{1}{k!} x^x$, the $x^x$ term of $(x+1)^x$ is approximately
$$ x^x (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{x!} ) \approx x^x e$$
as $x \to \infty$. Thus, the $x^{x+1}$ term of $x(x+1)^x$ is $ex^{x+1}$. So the $x^{x+1}$ powers in the numerator cancel.
Now we need to find the $x^x$ powers in the numerator. The first summand $ex^{x+1}$ doesn't provide any, so let's look at the second summand $x(x+1)^x$. Go back to the expansion of $(x+1)^x$:
$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$ 
Now we are looking for all terms with $x^{x-1}$ powers. Since the $x^k$ term of $\binom{x}{k}$ is approximately 
$$ \frac{-1-2-\cdots -k}{k!} = -\frac{k(k-1)}{2 k!} = -\frac{1}{2} \frac{1}{(k-2)!}$$
we have that the $x^{x-1}$ term of $(x+1)^x$ is approximately
$$ x^{x-1} \cdot (-\frac{1}{2}) (0 + 0 + 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots) \approx x^{x-1} (-\frac{e}{2}) $$
as $x \to \infty$. Thus, the $x^{x}$ term in the numerator is
$$ 0 + x \cdot (x^{x-1} (-\frac{e}{2} )) = \boxed{\frac{e}{2} x^{x}} .$$
Now consider the denominator. We already know that the $x^x$ term of $(x+1)^x$ tends to $e x^x$ as $x \to \infty$.
So, the limit becomes
\begin{align*}
\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} &= \lim_{x \to \infty} \frac{\frac{e}{2} x^x }{e x^x} \\
&= \frac{1}{2}.
\end{align*}
No elegant tricks. Just Taylor expansions and handwaving.
