# is equal to : Option 1) Option 2) Option 3) Option 4)

Answers (1)
P Prateek Shrivastava

As we learnt in

Method of Rationalisation -

Rationalisation method is used when we have RADICAL SIGNS in an expression.(like  1/2,  1/3 etc) and there exists a negative sign between two terms of an algebraic expression.

- wherein

$\lim_{x\rightarrow a}\:\frac{x-a}{\sqrt{x}-\sqrt{a}}$

$\therefore \:\frac{(x-a)(\sqrt{x}+\sqrt{a})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}$

$=\sqrt{x}+\sqrt{a}$

$=\sqrt{a}+\sqrt{a}$

$=2\sqrt{a}$

$\lim_{x\rightarrow 3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$

$\Rightarrow \lim_{x\rightarrow 3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}\times \frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{2x-4}+\sqrt{2}}\times \frac{\sqrt{3x}+3}{\sqrt{3x}+3}$

$\Rightarrow \lim_{x\rightarrow 3}\frac{3x-9}{2x-4-2}\times \frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{3x}+3}$

$\Rightarrow \lim_{x\rightarrow 3}\frac{3\left ( x-3 \right )}{2\left ( x-3 \right )}\times \frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{3x}+3}$

$\frac{3}{2}\cdot \frac{\sqrt{2\times 3-4}+\sqrt{2}}{\sqrt{3\times 3}+3}$

$\frac{3}{2}\times \frac{\sqrt{2}+\sqrt{2}}{6}$

$\Rightarrow \frac{2\sqrt{2}}{4}= \frac{\sqrt{2}}{2}= \frac{1}{\sqrt{2}}$

Option 1)

This option is incorrect.

Option 2)

This option is correct.

Option 3)

This option is incorrect.

Option 4)

This option is incorrect.

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