Question

Prove:

$\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sqrt{7+x}-\sqrt{7+x^2}}{\sqrt{7-x^2}-\sqrt{7-x}}=1$

Answer 1

Given,

${lim}_{x\to 0}\frac{\sqrt{7+x}-\sqrt{7+x^2}}{\sqrt{7-x^2}-\sqrt{7-x}}$

Apply L-hospital rule,

${lim}_{x\to 0}\frac{\frac{d}{dx}(\sqrt{7+x}-\sqrt{7+x^2})}{\frac{d}{dx}(\sqrt{7-x^2}-\sqrt{7-x})}$

${lim}_{x\to 0}\frac{\frac{1}{2\sqrt{7+x}}-\frac{1}{2\sqrt{7+x^2}}\times (0+2x)}{\frac{1}{2\sqrt{7-x^2}}\times (0-2x)-\frac{1}{2\sqrt{7-x}}\times (0-1)}$

${lim}_{x\to 0}\frac{\frac{1}{2\sqrt{7+x}}-\frac{2x}{2\sqrt{7+x^2}}}{\frac{2x}{2\sqrt{7-x^2}}+\frac{1}{2\sqrt{7-x}}}$

$=\frac{\frac{1}{2\sqrt{7+0}}-0}{0+\frac{1}{2\sqrt{7-0}}}$

$=\frac{\frac{1}{2\sqrt{7}}}{\frac{1}{2\sqrt{7}}}$

$=1$

Hence, proved.