Question

The value of $\underset{x\to 2}{\lim }\frac{5}{\sqrt{2}-\sqrt{X}}$ is

• A. $10\sqrt{2}$
• B. $+\infty$
• C. $-\infty$
• D. does not exist.

Answer 1

The correct option is D

does not exist.

Explanation of the correct option.

Step $1$: Compute L.H.L.

$\underset{x\to 2}{\lim }\frac{5}{\sqrt{2}-\sqrt{X}}$

$=\underset{h\to 0}{\lim }\frac{5}{\sqrt{2}-\sqrt{2-h}}$

$=\underset{h\to 0}{\lim }\frac{5}{\sqrt{2}-\sqrt{2-h}}\times \frac{\sqrt{2}+\sqrt{2-h}}{\sqrt{2}+\sqrt{2-h}}$

$=\underset{h\to 0}{\lim }\frac{5\left(\sqrt{2}+\sqrt{2-h}\right)}{2-2+h}$

$=+\infty$

Step $2$: Compute R.H.L.

$\underset{x\to 2}{\lim }\frac{5}{\sqrt{2}-\sqrt{X}}$

$=\underset{h\to 0}{\lim }\frac{5}{\sqrt{2}-\sqrt{2+h}}$

$=\underset{h\to 0}{\lim }\frac{5}{\sqrt{2}-\sqrt{2-h}}\times \frac{\sqrt{2}+\sqrt{2+h}}{\sqrt{2}+\sqrt{2+h}}$

$=\underset{h\to 0}{\lim }\frac{5\left(\sqrt{2}+\sqrt{2-h}\right)}{2-2-h}$

$=-\infty$

Since, L.H.L.$\ne$R.H.L.

Therefore limit does not exist.

Hence, option $\left(D\right)$ is the correct option.