Question:
\(\displaystyle\lim_{x \to 3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}}\) is equal to :
\(\displaystyle\lim_{x \to 3} \frac{\sqrt{3x} - 3}{\sqrt{2x - 4} - \sqrt{2}}\) is equal to :
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The limit is of the form 0/0; rationalise numerator and denominator to cancel the factor \((x - 3)\).
Updated On: Nov 11, 2026
- \(\sqrt{3}\)
- \(\frac{1}{\sqrt{2}}\)
- \(\frac{\sqrt{3}}{2}\)
- \(\frac{1}{2\sqrt{2}}\)
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The Correct Option is B
Solution and Explanation
At \(x = 3\), numerator \(\sqrt{9} - 3 = 0\) and denominator \(\sqrt{2} - \sqrt{2} = 0\), so it is a 0/0 form.
\(\sqrt{3x} - 3 = \frac{3x - 9}{\sqrt{3x} + 3} = \frac{3(x - 3)}{\sqrt{3x} + 3}\).
\(\sqrt{2x - 4} - \sqrt{2} = \frac{2x - 6}{\sqrt{2x-4} + \sqrt{2}} = \frac{2(x - 3)}{\sqrt{2x-4} + \sqrt{2}}\).
Limit \(= \lim_{x\to 3} \frac{3}{2}\cdot\frac{\sqrt{2x-4} + \sqrt{2}}{\sqrt{3x} + 3} = \frac{3}{2}\cdot\frac{2\sqrt{2}}{6} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\).
Answer: (2) \(\frac{1}{\sqrt{2}}\)
\(\sqrt{3x} - 3 = \frac{3x - 9}{\sqrt{3x} + 3} = \frac{3(x - 3)}{\sqrt{3x} + 3}\).
\(\sqrt{2x - 4} - \sqrt{2} = \frac{2x - 6}{\sqrt{2x-4} + \sqrt{2}} = \frac{2(x - 3)}{\sqrt{2x-4} + \sqrt{2}}\).
Limit \(= \lim_{x\to 3} \frac{3}{2}\cdot\frac{\sqrt{2x-4} + \sqrt{2}}{\sqrt{3x} + 3} = \frac{3}{2}\cdot\frac{2\sqrt{2}}{6} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\).
Answer: (2) \(\frac{1}{\sqrt{2}}\)
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