I copied the OP's work from the image site.
Eventually, the image will be taken down. To preserve the meaning in this (eventually archived) thread, here's what that image contains:
6. Calculate the indicated limits: (a) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\,\) (b) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
My working:
(a) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{x^2\, +\, 3x}{x^2\, -\, 9}\)
\(\displaystyle \dfrac{(-3)^2\, +\, 3(-3)}{(-3)^2\, -\, 9}\, =\, \dfrac{9\, -\, 9}{9\, -\, 9}\, =\, \dfrac{0}{0}\)
\(\displaystyle \dfrac{x\, (x\, +\, 3)}{(x\, -\, 3)\,(x\, +\, 3)}\, =\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle \dfrac{-3}{(-3)\, -\, 3}\, =\, \dfrac{-3}{-6}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, -3}\, \)\(\displaystyle =\, \dfrac{1}{2}\)
(b) \(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
\(\displaystyle \dfrac{4\, -\, (2)^2}{(2)^2\, -\, (2)\, -\, 2}\, =\, \dfrac{4\, -\, 4}{4\, -\, 2\, -\, 2}\, =\, \dfrac{0}{0}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{4\, -\, x^2}{x^2\, -\, x\, -\, 2}\)
\(\displaystyle \dfrac{(2\, +\, x)\, (2\, -\, x)}{(x\, +\, 1)\, (x\, -\, 2)}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\, \)\(\displaystyle \dfrac{(2\, +\, 2)\, (2\, -\, 2)}{(2\, +\, 1)\, (2\, -\, 2)}\, =\, \dfrac{4}{3}\)
\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 2}\,\)\(\displaystyle =\, \dfrac{4}{3}\)