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!

Consider $$y=\frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}=\frac{ex^{x+1}}{(x+1)^x}-x$$ Now
$$z=\frac{ex^{x+1}}{(x+1)^x}\implies \log(z)=1+(x+1)\log(x)-x\log(x+1)=1+\log(x)-x \log\left(1+\frac 1x\right)$$ Now, using Taylor $$\log\left(1+\frac 1x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right)$$ makes $$\log(z)=\log \left({x}\right)+\frac{1}{2 x}-\frac{1}{3
   x^2}+O\left(\frac{1}{x^3}\right)$$ $$z=e^{\log(z)}=x+\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ which finally makes $$y=\frac{1}{2}-\frac{5}{24 x}+O\left(\frac{1}{x^2}\right)$$ showing the limit and how it is approached.
Using $x=10$ which is really small, the exact value would be $$\frac{100000000000 e}{25937424601}-10\approx 0.480153$$ while the above approximation would simply give $\frac{23}{48}\approx 0.479167$.
