Answer
$\lim\limits_{x \to \ 3}\frac{x^2-9}{2x-6}=\frac{x+3}{2}=\frac{3+3}{2}=3$
Work Step by Step
If we want to calculate the value of $\lim\limits_{x \to \ 3}\frac{x^2-9}{2x-6}$,
Here, we cannot use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, because it would mean that:
$\lim\limits_{x \to \ 3}\frac{x^2-9}{2x-6}=\frac{(3)^2-9}{2(3)-6}=\frac{0}{0}$ which cannot be a limit.
Therefore, we have to simplify the fraction:
$\frac{x^2-9}{2x-6}=\frac{(x-3)(x+3)}{2(x-3)}=\frac{x+3}{2}$
If we substitute this function into the limit, we get:
$\lim\limits_{x \to \ 3}\frac{x^2-9}{2x-6}=\frac{x+3}{2}=\frac{3+3}{2}=3$
