# Limit of a quotient with denominator approaching zero: $\lim_{x\to -1}\frac{\sqrt {x ^ 2 + 8} - 3}{x + 1}$

I was asked to find the limit of the following:

$$\lim_{x\to -1}\frac{\sqrt {x ^ 2 + 8} - 3}{x + 1}$$

I have tried using the Limit Laws but am always getting $\frac {0}{0}$. The answer given is $-\frac {1}{3}$. Can someone give me some hints to arriving at the answer?

$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} \cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}.$$
• @ALGEAN Thanks for the edit! Those curly brackets$\ldots$ Feb 2, 2014 at 9:42
$$\lim_{x\to -1}\frac{(\sqrt {x ^ 2 + 8} - 3)(\sqrt {x ^ 2 + 8} + 3)}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)}$$ $$\lim_{x\to -1}\frac{x^2+8-9}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)}$$ $$\lim_{x\to -1}\frac{(x+1)(x-1)}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)}$$ $$=\frac{-2}{6}=\frac{-1}{3}$$
$$\lim_{x\to -1}\frac{\sqrt {x ^ 2 + 8} - 3}{x + 1} = \lim_{x\to -1} \frac{\sqrt {x ^ 2 + 8} - 3}{x + 1} \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3} = \lim_{x\to -1} \frac{x^2+8-9}{(x+1)(\sqrt{x^2+8}+3)}.$$ Now simplify $\frac{x^2-1}{x+1}$ and compute the limit.