Question

Evaluate the following limits: ${\displaystyle \underset{x\to 2}{lim}\left[\frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}}\right]}$

Answer 1

${\displaystyle \underset{x\to 2}{lim}\left[\frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}}\right]}$

= ${\displaystyle \underset{x\to 2}{lim}[\frac{x^3-8}{\sqrt{x+2}-\sqrt{3x-2}}\times \frac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{x+2}+\sqrt{3x-2}}]}$

= ${\displaystyle \underset{x\to 2}{lim}\frac{(x^3-8)(\sqrt{x+2}+\sqrt{3x-2})}{(x+2)-(3x-2)}}$

= ${\displaystyle \underset{x\to 2}{lim}\frac{(x^3-8)(\sqrt{x+2}+\sqrt{3x-2})}{-2x+4}}$

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

= ${\displaystyle \underset{x\to 2}{lim}\frac{(x^2+2x+4)(\sqrt{x+2}+\sqrt{3x-2})}{-2} ...\left[\begin{array}{c}\text{As} x\to 2\text{,} x\ne 2\\ \therefore x-2\ne \end{array}\right]}$

= ${\displaystyle -\frac{1}{2}\underset{x\to 2}{lim}(x^2+2x+4)\times \underset{x\to 2}{lim}(\sqrt{x+2}+\sqrt{3x-2})}$

= ${\displaystyle -\frac{1}{2}\times [{\left(2\right)}^2+2\left(2\right)+4]\times (\sqrt{2+2}+\sqrt{3\left(2\right)-2})}$

= ${\displaystyle -\frac{1}{2}\times 12\times (2+2)}$

= – 6 x 4
= – 24