What is the value of $\lim_{x \to 0} \frac{\ln(x + 1)}{x^2 + x}$?

Core Concept

Correct. Direct substitution yields the indeterminate form $0/0$. Applying L'Hôpital's Rule by differentiating the numerator and denominator gives $\frac{1/(x+1)}{2x+1}$, which evaluates to $1/1 = 1$ at $x=0$.