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!

Change $x=\frac{1}{t}$ and simplify:
$$L=\lim_{t\to 0} \frac{e-(1+t)^{1/t}}{t(1+t)^{1/t}}=L'H=$$
$$\lim_{t\to 0} \frac{(1+t)\ln{(1+t)}-t}{t^2(1+t)+t^2-t(1+t)\ln{(1+t)}}=Taylor=$$
$$\lim_{t\to 0} \frac{\frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=\frac{1}{2}.$$
