\lim_{x\rightarrow 3}\tfrac{\sqrt{3x}-3}{\sqrt{2x-4}- \sqrt{2}}        is equal to :

  • Option 1)

    \sqrt{3}

  • Option 2)

    \frac{1}{\sqrt{2}}

  • Option 3)

    \frac{\sqrt{3}}{2}

  • Option 4)

    \frac{1}{2\sqrt{2}}

 

Answers (1)

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)

\sqrt{3}

This option is incorrect.

Option 2)

\frac{1}{\sqrt{2}}

This option is correct.

Option 3)

\frac{\sqrt{3}}{2}

This option is incorrect.

Option 4)

\frac{1}{2\sqrt{2}}

This option is incorrect.

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