JEE Main On 8 April 2017 Question 16

Question: $ \underset{x\to 3}{\mathop{\lim }},\frac{\sqrt{3x}-3}{\sqrt{2x}-4-\sqrt{2}} $ is equal to: [JEE Online 08-04-2017]

Options:

A) $ \frac{1}{\sqrt{2}} $

B) $ \frac{1}{2\sqrt{2}} $

C) $ \frac{\sqrt{3}}{2} $

D) $ \sqrt{3} $

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Answer:

Correct Answer: A

Solution:

$ \underset{x\to 3}{\mathop{\lim }},\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} $

Rationalize $ \underset{x\to 3}{\mathop{\lim }},\frac{(3x-9)\times ( \sqrt{2x-4}+\sqrt{2} )}{{( 2x-4 )-2}\times ( \sqrt{3x}+3 )} $

$ =\underset{x\to 3}{\mathop{\lim }},\frac{( 3x-3 )}{2( x-3 )}\times \frac{\sqrt{2x-4}+\sqrt{2}}{( \sqrt{3}x+3 )} $

$ =\frac{3}{2}\times \frac{2\sqrt{2}}{6}=\frac{1}{\sqrt{2}} $