Answer
$\lim\limits_{h \to 0}\frac{(3+h)^{2}-3^{2}}{h}=6$
Work Step by Step
Numerical Estimation:
for $h=-0.1$: $\frac{(3-0.1)^{2}-3^{2}}{-0.1}\approx5.9$
for $h=-0.01$: $\frac{(3-0.01)^{2}-3^{2}}{-0.01}\approx5.99$
for $h=-0.001$: $\frac{(3-0.001)^{2}-3^{2}}{-0.001}\approx5.999$
for $h=-0.0001$: $\frac{(3-0.0001)^{2}-3^{2}}{-0.0001}\approx5.9999$
$\lim\limits_{h \to 0^{-}}\frac{(3+h)^{2}-3^{2}}{h}\approx6$
Numerical Estimation:
for $h=0.1$: $\frac{(3+0.1)^{2}-3^{2}}{0.1}\approx6.1$
for $h=0.01$: $\frac{(3+0.01)^{2}-3^{2}}{0.01}\approx6.01$
for $h=0.001$: $\frac{(3+0.001)^{2}-3^{2}}{0.001}\approx6.001$
for $h=0.0001$: $\frac{(3+0.0001)^{2}-3^{2}}{0.0001}\approx6.0001$
$\lim\limits_{h \to 0^{+}}\frac{(3+h)^{2}-3^{2}}{h}\approx6$
