EXAMPLE $5$ Evaluate the limit below.

$\underset{h6}{lim}\frac{(7+h{)}^2-49}{h}$

SOLUTION If we define

$F(h)=((7+h{)}^2-h)$

then we can't compute the limit by letting $h=0$ since $F(0)$ is undefined, but If we simplify $F(h)$ algebraically, we find that

$F(h)=\frac{(ubrace(\frac{-}{h}\text{u}\text{b}\text{r}\text{a}\text{c}\text{e}))}{h}=$

$($Recall that we consider only $h=0$ when letting $h$ approach $0.)$ Thus

$\underset{h0}{lim}\frac{(7+h{)}^2-49}{h}=\underset{h0}{lim}(14+h)=$

$$