Answer
$ f(3)=\frac{9-3^{2}}{3-3}=\frac{0}{0}$
The function has the indeterminate form $\frac{0}{0}$ at h=3.
Transforming algebraically and canceling, we have
$\frac{9-h^{2}}{h-3}=\frac{-(h+3)(h-3)}{h-3}=-h-3$
Evaluating using continuity, we get
$\lim\limits_{h \to 3}\frac{9-h^{2}}{h-3}=\lim\limits_{h \to 3}(-h-3)=-3-3=-6$
Work Step by Step
See the answer above.
