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$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ is equal to :

JEE · Math · previous-year question

  1. A.$$\sqrt 3 $$
  2. B.$${1 \over {\sqrt 2 }}$$correct
  3. C.$${{\sqrt 3 } \over 2}$$
  4. D.$${1 \over {2\sqrt 2 }}$$

Answer

B. $${1 \over {\sqrt 2 }}$$

Explanation

Given, $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ Here if you put x = 3 in $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$$ you will get $${0 \over 0}$$ form. So, we can apply L' Hospital rule $$\therefore\,\,\,$$ $$\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ = $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$$ (applying L' Hospital rule) = $${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$$ = $${1 \over 2}$$ $$ \times $$ $$\sqrt 2 $$ = $${1 \over {\sqrt 2 }}$$

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