Answer
$\lim\limits_{x \to 2}\frac{\sqrt (x^{2}+12) -4}{x-2} =\frac{1}{2} = 0.5$
Work Step by Step
$\frac{\sqrt (x^{2}+12) -4}{x-2} = \frac{(\sqrt (x^{2}+12) -4)(\sqrt (x^{2}+12) +4)}{(x-2)(\sqrt (x^{2}+12) +4)} = \frac{x^{2}+12-16}{(x-2)(\sqrt (x^{2}+12) +4)} = \frac{x^{2}-4}{(x-2)(\sqrt (x^{2}+12) +4)}$
= $\frac{(x+2)(x-2)}{(x-2)(\sqrt (x^{2}+12) +4)}$
= $\frac{(x+2)}{(\sqrt (x^{2}+12) +4)}$
Now,
$\lim\limits_{x \to 2}\frac{\sqrt (x^{2}+12) -4}{x-2} = \lim\limits_{x \to 2}\frac{(x+2)}{(\sqrt (x^{2}+12) +4)} = \frac{2+2}{4+4} = \frac{4}{8} = \frac{1}{2} = 0.5$
