Evaluating a 0/0 limit using L'Hopital's Rule

Substituting $x = 4$ gives $\sqrt{9} - 3 = 0$ in the numerator and $4 - 4 = 0$ in the denominator, so the limit has the indeterminate form $0/0$. This confirms L'Hopital's Rule applies.

Differentiate numerator and denominator separately: the derivative of $\sqrt{x+5} - 3$ is $\frac{1}{2\sqrt{x+5}}$, and the derivative of $x - 4$ is $1$. So the limit becomes $\lim_{x\to 4} \frac{1}{2\sqrt{x+5}}$.

Substituting $x = 4$ into this new expression gives $\frac{1}{2\sqrt{9}} = \frac{1}{6}$, which is no longer indeterminate. Therefore the original limit equals $\frac{1}{6}$.