Mathematics

This is a derivative in disguise—you're finding the instantaneous rate of change of $\sqrt{x}$ at $x=16$. The key insight is that direct substitution gives $\frac{0}{0}$, so you need to algebraically eliminate the $h$ from the denominator. Multiply by the conjugate $\sqrt{16+h}+4$ to rationalize the numerator and reveal what $h$ cancels with.

This is a derivative in disguise—you're finding the instantaneous rate of change of $\sqrt{x}$ at $x=16$. The key insight is that direct substitution gives $\frac{0}{0}$, so you need to algebraically eliminate the $h$ from the denominator. Multiply by the conjugate $\sqrt{16+h}+4$ to rationalize the numerator and reveal what $h$ cancels with.