Solving $\lim_{x\to 0} \frac{e^x-1-x}{x^2}$ without L'Hôpital's Rule.

Question

I want to solve this limit without the use of L'Hôpital's rule:

$$\lim_{x\to 0} \frac{e^x-1-x}{x^2}.$$

Answer 1

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