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

Answer: (D)

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

$LHL=\underset{x\to 2^{-}}{\lim }\frac{5}{\sqrt{2}-\sqrt{x}}$

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

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

$RHL={\lim }_{x\to 2^{+}}\frac{5}{\sqrt{2}-\sqrt{x}}$

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

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

$\because LHL\ne RHL$

$\therefore$ Limit does not exist