JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 24)
$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ is equal to :
$$\sqrt 3 $$
$${1 \over {\sqrt 2 }}$$
$${{\sqrt 3 } \over 2}$$
$${1 \over {2\sqrt 2 }}$$
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)
= $${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$$
= $${1 \over 2}$$ $$ \times $$ $$\sqrt 2 $$
= $${1 \over {\sqrt 2 }}$$
$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$
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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