Answer to Question #239397 in Calculus for sam

Question

Evaluate the limit $\displaystyle{\lim\limits_{x \to 0} \dfrac{e^{x}-x-1}{x^2}}$ by using l'Hopital's Rule twice.

Answer 1

$\lim\limits_{x\rightarrow0} \frac{e^x-x-1}{x^2}$

This is a $\frac{0}{0}$ form, so applying L'Hopital's rule here, we get:

$=\lim\limits_{x\rightarrow0} \frac{e^x-1}{2x}$

Again, this is a $\frac{0}{0}$ form, so applying L'Hopital's rule here, we get:

$=\lim\limits_{x\rightarrow0} \frac{e^x}{2}$

$=\frac{e^0}{2}\\ =\frac{1}{2}$