Answer
$\lim\limits_{x \to -1}\frac{\sqrt (x^{2}+8) - 3}{x+1} = -\frac{1}{3} = 0.33$
Work Step by Step
$\frac{\sqrt (x^{2}+8) - 3}{x+1}$
=$\frac{(\sqrt (x^{2}+8) - 3)(\sqrt (x^{2}+8) + 3)}{(x+1)(\sqrt (x^{2}+8) + 3)} $
= $\frac{(x^{2}+8)-9}{(x+1)(\sqrt (x^{2}+8) + 3)}$
= $\frac{(x^{2}-1)}{(x+1)(\sqrt (x^{2}+8) + 3)}$
= $\frac{(x-1)(x+1)}{(x+1)(\sqrt (x^{2}+8) + 3)} = \frac{(x-1)}{(\sqrt (x^{2}+8) + 3)} $
Now,
$\lim\limits_{x \to -1}\frac{\sqrt (x^{2}+8) - 3}{x+1} =\lim\limits_{x \to -1} \frac{(x-1)}{(\sqrt (x^{2}+8) + 3)} = \frac{-1-1}{3+3} =-\frac{2}{6} = -\frac{1}{3} = -0.33$
