<s>[INST] Evaluate $\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$ $$\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$$\lim_{x\to 0}\frac{xe^x+x-2e^x+2}{x^3}$$
Limit is in the form "$\frac{0}{0}$", using L'hopital once:
$$\lim_{x\to 0}\frac{e^x+xe^x+1-2e^x}{3x^2}=\lim_{x\to 0}\frac{-e^x+xe^x+1}{3x^2}$$
Again the limit is of the form "$\frac{0}{0}$", using L'hopital again:
$$\lim_{x\to 0}\frac{-e^x+e^x+xe^x}{6x}=\lim_{x\to 0}\frac{xe^x}{6x}$$
Once again the limit is of the form "$\frac{0}{0}$", using L'hopital one more time:
$$\lim_{x\to 0}\frac{e^x+xe^x}{6}=\frac{1}{6}$$
Is there a way to solve it without L'hopital?
 [/INST] series expansion of $\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $
around $x=0$
$$\lim_{x\rightarrow 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$
$\displaystyle \lim_{x\rightarrow 0}\frac{x\bigg(\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\bigg)+1\bigg)-2\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots -1\bigg)}{x^3}$
$$\lim_{x\rightarrow 0}\frac{x^3(\frac{1}{2!}-\frac{2}{3!})+x^4()+\cdots}{x^3}=\frac{1}{6}$$
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