Limit of $x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$ when $x\to+\infty$ 
Find $$\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$$

I calculated $\lim\limits_{x\to\infty}\left(\frac{x+2}{x+1}\right)^x=e$ but then the limit in question becomes $0\times \infty $ form, and further solution becomes messy.

Please tell a solution without the use of series expansions because I have no knowledge of these.
 A: \begin{align}
\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)&=\lim_{x\to\infty}\frac{e-\left(\frac{x+2}{x+1}\right)^x}{\frac1x}\\
&=\lim_{x\to\infty}\frac{\frac{\left(\frac{1}{x+1}+1\right)^x \left(x-(x+1) (x+2) \log
   \left(\frac{1}{x+1}+1\right)\right)}{(x+1) (x+2)}}{-\frac1{x^2}}\\
&=-e\lim_{x\to\infty}\left(x-(x+1) (x+2) \log
   \left(\frac{1}{x+1}+1\right)\right)\\
&=-e\lim_{x\to0}\frac{x-(x+1) (2 x+1) \left(x-\frac{3 x^2}{2}+\frac{7 x^3}{3}-\frac{15 x^4}{4}+o(x^4)\right)}{x^2}\\
&=-e \lim_{x\to 0}(\frac{15 x^4}{2}+\frac{79
   x^3}{12}-\frac{x^2}{4}+\frac{x}{6}-\frac{3}{2}+o(x^2))\\
&=\frac32e
\end{align}
A: Rewriting $\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x$ with Laurent series expansion rather than calculating out the limit you get:
$$\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x=e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some higher order terms}$$
Thus,
$$x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=xe-x\left(e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some h.o.t.}\right)=\frac{3e}{2} +\frac{83e}{24x} +\text{some h.o.t.}$$
hence
$$
\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=\frac{3e}{2}$$
A: Without Taylor series: change the variable $t=\frac{1}{x+1}$ to get
$$
x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=
\frac{1-t}{t}\left(e-(1+t)^{\frac{1}{t}-1}\right)=
\frac{1-t}{1+t}\cdot \frac{e(1+t)-(1+t)^{1/t}}{t}=\\
=\underbrace{\frac{1-t}{1+t}}_{\to 1}e+\underbrace{\frac{1-t}{1+t}}_{\to 1}e\cdot\frac{1-e^{\frac{1}{t}\ln(1+t)-1}}{t}.
$$
The first term goes trivially to $1$ as $t\to 0$. Let's calculate the second limit. Denote $s(t)=\frac{1}{t}\ln(1+t)-1$. We have
$$
\frac{1-e^{s(t)}}{t}=\frac{1-e^{s(t)}}{s(t)}\cdot\frac{s(t)}{t}=
\underbrace{\frac{1-e^{s(t)}}{s(t)}}_{I}\cdot\underbrace{\frac{\ln(1+t)-t}{t^2}}_{II}.
$$
The limits $I$ and $II$ are calculated by L'Hospital:
\begin{eqnarray}
\lim_{t\to 0}s(t)&=&\lim_{t\to 0}\frac{\ln(1+t)}{t}-1=\lim_{t\to 0}\frac{\frac{1}{1+t}}{1}-1=0,\\
\lim I&=&\lim_{s\to 0}\frac{1-e^s}{s}=\lim_{s\to 0}\frac{-e^s}{1}=-1.\\
\lim II&=&\lim_{t\to 0}\frac{\ln(1+t)-t}{t^2}=\lim_{t\to 0}\frac{\frac{1}{1+t}-1}{2t}=\lim_{t\to 0}\frac{-t}{2t(1+t)}=-\frac{1}{2}.
\end{eqnarray}
Finally, the limit is $e+e\cdot (-1)\cdot\frac{-1}{2}=\frac32 e$.
