Question

Find :
$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2}\right)}{x^2-10}$

Answer 1

Apply L'Hospitals rule
i.e. differentiate both numerator and denominator separately
$\underset{x\to \sqrt{10}}{lim}\frac{1}{\sqrt{7+2x}\left(2x\right)}$
$\frac{1}{2\sqrt{10}\sqrt{7+2\sqrt{10}}}=\frac{1}{2\sqrt{10}}\times \frac{1}{\sqrt{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{2}\right)}^2+2.\sqrt{5}\sqrt{2}}}$
$\frac{1}{2\sqrt{10}}\times \frac{1}{\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^2}}=\frac{1}{2\sqrt{10}\left(\sqrt{5}+\sqrt{2}\right)}=\frac{\sqrt{5}-\sqrt{2}}{6\sqrt{10}}$