<s>[INST] 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.
 [/INST] 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
$$
\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=\frac{3e}{2}$$
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