Question

Evaluate the following limits: ${\displaystyle \underset{\text{v}\to \sqrt{2}}{lim}\left[\frac{{\text{v}}^2+\text{v}\sqrt{2}-4}{{\text{v}}^2-3\text{v}\sqrt{2}+4}\right]}$

Answer 1

${\displaystyle \underset{\text{v}\to \sqrt{2}}{lim}\left[\frac{{\text{v}}^2+\text{v}\sqrt{2}-4}{{\text{v}}^2-3\text{v}\sqrt{2}+4}\right]}$

Consider, ${\displaystyle {\text{v}}^2+\text{v}\sqrt{2}-4={\text{v}}^2+\sqrt{2}\text{v}-4}$

= ${\displaystyle {\text{v}}^2+2\sqrt{2}\text{v}-\sqrt{2}\text{v}-4}$

= ${\displaystyle \text{v}(\text{v}+2\sqrt{2})-\sqrt{2}(\text{v}+2\sqrt{2})}$

= ${\displaystyle (\text{v}+2\sqrt{2})(\text{v}-\sqrt{2})}$

${\displaystyle {\text{v}}^2-3\text{v}\sqrt{2}+4={\text{v}}^2-3\sqrt{2}\text{v}+4}$

= ${\displaystyle {\text{v}}^2-2\sqrt{2}\text{v}-\sqrt{2}\text{v}+4}$

= ${\displaystyle \text{v}(\text{v}-2\sqrt{2})-\sqrt{2}(\text{v}-2\sqrt{2})}$

= ${\displaystyle (\text{v}-2\sqrt{2})(\text{v}-\sqrt{2})}$

∴ ${\displaystyle \underset{\text{v}\to \sqrt{2}}{lim}\left[\frac{{\text{v}}^2+\text{v}\sqrt{2}-4}{{\text{v}}^2-3\text{v}\sqrt{2}+4}\right]}$

= ${\displaystyle \underset{\text{v}\to \sqrt{2}}{lim}\frac{(\text{v}+2\sqrt{2})(\text{v}-\sqrt{2})}{(\text{v}-2\sqrt{2})(\text{v}-\sqrt{2})}}$

= ${\displaystyle \underset{\text{v}\to \sqrt{2}}{lim}\frac{\text{v}+2\sqrt{2}}{\text{v}-2\sqrt{2}} ...\left[\begin{array}{c}\text{As} \text{v}\to \sqrt{2}\text{,}\text{v}\ne \sqrt{2}\\ \therefore \text{v}-\sqrt{2}\ne 0\end{array}\right]}$

= ${\displaystyle \frac{\sqrt{2}+2\sqrt{2}}{\sqrt{2}-2\sqrt{2}}}$

= ${\displaystyle \frac{3\sqrt{2}}{-\sqrt{2}}}$

= – 3