MCQ
$\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x}  - 3}}{{\sqrt {2x - 4}  - \sqrt 2 }}$ is equal to
  • A
    $\sqrt 3 $
  • $\frac{1}{{\sqrt 2 }}$
  • C
    $\frac{{\sqrt 3 }}{2}$
  • D
    $\frac{1}{{2\sqrt 2 }}$

Answer

Correct option: B.
$\frac{1}{{\sqrt 2 }}$
b
Let $A = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x}  - 3}}{{\sqrt {2x - 4}  - \sqrt 2 }}$

Rationalise 

$ \Rightarrow A = \,\,\,\mathop {\lim }\limits_{x \to 3} \frac{{\left( {3x - 9} \right) \times \left( {2x - 4 + \sqrt 2 } \right)}}{{\left\{ {\left( {2x - 4 - 2} \right)} \right\} \times \left( {\sqrt {3x}  + 3} \right)}}$

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

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

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