Question

$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}$ is equal to -

• A. $\frac{1}{\sqrt{40}(\sqrt{5}+\sqrt{2})}$
• B. $\frac{-1}{\sqrt{10}[\sqrt{7+2\sqrt{10}}+\sqrt{5}+2]}$
• C. $1$
• D. $\frac{1}{\sqrt{10}\left(\sqrt{5}+\sqrt{2}\right)}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{40}(\sqrt{5}+\sqrt{2})}$

$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2}\right)}{x^2-10}$
Apply L'Hospital Rule
$\begin{array}{c}\underset{x\to \sqrt{10}}{lim}\frac{\left(\frac{2}{2\sqrt{7+2x}}\right)}{2x}\\ =\underset{x\to \sqrt{10}}{lim}\frac{1}{2x\sqrt{7+2x}}\\ =\frac{1}{2\sqrt{10}\sqrt{7+2\sqrt{10}}}\\ =\frac{1}{2\sqrt{10}}\times \frac{1}{\sqrt{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{5}\sqrt{2}}}\\ =\frac{1}{2\sqrt{10}}\times \frac{1}{\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^2}}\\ =\frac{1}{2\sqrt{10}}\times \frac{1}{\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^2}}\\ =\frac{1}{\sqrt{40}\left(\sqrt{5}+\sqrt{2}\right)}\end{array}$