Question

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

Answer 1

1/2

Answer 2

The limit is in the indeterminate form $\frac{0}{0}$ as $x \to 0$. Using L'Hôpital's Rule, we differentiate the numerator and the denominator. After the first application, we get $\lim_{x \to 0}\frac{e^x-1}{2x}$, which is still $\frac{0}{0}$. Applying L'Hôpital's Rule again, we get $\lim_{x \to 0}\frac{e^x}{2}$. Evaluating this limit gives $\frac{e^0}{2} = \frac{1}{2}$.

Alternatively, using the Taylor series expansion $e^x = 1 + x + \frac{x^2}{2!} + \dots$, the expression simplifies to $\frac{1}{2} + \frac{x}{6} + \dots$, and the limit as $x \to 0$ is $\frac{1}{2}$.