Question

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

Answer 1

$-\frac{1}{2}$

Answer 2

The limit is of the form $\frac{0}{0}$ as $x \to 0$. Applying L'Hopital's Rule twice:

$\lim_{x \to 0} \frac{\ln(1+x)-x}{x^2} = \lim_{x \to 0} \frac{\frac{1}{1+x}-1}{2x} = \lim_{x \to 0} \frac{-\frac{1}{(1+x)^2}}{2} = -\frac{1}{2}$.

Alternatively, using the Taylor expansion $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$, the expression becomes $\frac{(x - \frac{x^2}{2} + O(x^3)) - x}{x^2} = \frac{-\frac{x^2}{2} + O(x^3)}{x^2} = -\frac{1}{2} + O(x)$, which tends to $-\frac{1}{2}$ as $x \to 0$.