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

Question

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?

Answer 1

Hint:

Try to expand the fraction like this:

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

Answer 2

$$\lim_{x\to -1}\frac{(\sqrt {x ^ 2 + 8} - 3)(\sqrt {x ^ 2 + 8} + 3)}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)} $$ $$\lim_{x\to -1}\frac{x^2+8-9}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)} $$ $$\lim_{x\to -1}\frac{(x+1)(x-1)}{(x + 1)(\sqrt {x ^ 2 + 8} + 3)}$$ $$=\frac{-2}{6}=\frac{-1}{3}$$

Answer 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.