Question

${\displaystyle \underset{\text{x}\to -3}{lim}\frac{\sqrt{{\text{x}}^2+7-4}}{\text{x}+3}}$ is equal to ____________.

Options

• ${\displaystyle \frac{4}{3}}$

• ${\displaystyle -\frac{3}{4}}$

• None of these

• ${\displaystyle \frac{3}{4}}$

Answer 1

${\displaystyle \underset{\text{x}\to -3}{lim}\frac{\sqrt{{\text{x}}^2+7-4}}{\text{x}+3}}$ is equal to ${\displaystyle \underline{-\frac{3}{4}}}$.

Explanation:

${\displaystyle \underset{\text{x}\to -3}{lim}\frac{\sqrt{{\text{x}}^2+7-4}}{\text{x}+3}}$

${\displaystyle = \underset{\text{x}\to -3}{lim}\frac{{\text{x}}^2+7-16}{(\text{x}+3)({\text{x}}^2+7+4)}}$

${\displaystyle = \underset{\text{x}\to -3}{lim}\frac{{\text{x}}^2-9}{(\text{x}+3)\left(\sqrt{{\text{x}}^2+7+4}\right)}}$

${\displaystyle = \underset{\text{x}\to -3}{lim}\frac{\text{x}-3}{\left(\sqrt{{\text{x}}^2+7+4}\right)}}$

${\displaystyle =\frac{-3-3}{\left(\sqrt{9+7+4}\right)}}$

${\displaystyle =-\frac{3}{4}}$