Solving $\lim_{x\to 0} \frac{e^x-1-x}{x^2}$ without L'Hôpital's Rule. I want to solve this limit without the use of L'Hôpital's rule:
$$\lim_{x\to 0} \frac{e^x-1-x}{x^2}.$$

I would Taylor Expand 
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
and plug in to find
$$\lim_{x\to 0}\frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x}{x^2}=\lim_{x\to 0}\frac{\sum_{n=2}^\infty \frac{x^n}{n!}}{x^2}=\lim_{x\to 0}\frac{\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots}{x^2}=\lim_{x\to 0} \frac{1}{2}+\frac{1}{6}x+\frac{1}{24}x^2+\cdots=\frac{1}{2}.$$
