Evaluate $\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$

Question

$$\lim_{x\to 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$

$$\lim_{x\to 0}\frac{xe^x+x-2e^x+2}{x^3}$$

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?

Answer 1

series expansion of $\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $

around $x=0$

$$\lim_{x\rightarrow 0}\frac{x(e^x+1)-2(e^x-1)}{x^3}$$

$\displaystyle \lim_{x\rightarrow 0}\frac{x\bigg(\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\bigg)+1\bigg)-2\bigg(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots -1\bigg)}{x^3}$

$$\lim_{x\rightarrow 0}\frac{x^3(\frac{1}{2!}-\frac{2}{3!})+x^4()+\cdots}{x^3}=\frac{1}{6}$$

Answer 2

$$ \frac{x(e^x+1)-2(e^x-1)}{x^3}\approx\frac{x(2+x+x^2/2)-2x-x^2-x^3/3}{x^3}=\frac{x^3/2-x^3/3}{x^3}=\frac{1}{6}. $$