NIMCET 2019 Question Paper with solutions

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Question : 40 of 120

Marks: +1, -0

\lim↙{\x→3}{√{3x} - 3}/{√{2\x-4}-√2

is equal to

$√3$
$√3/2$
$1/{2√2$
$1/√2$

Solution:

Let

L=\lim↙{x→3}\;\;{√{3x}-3}/{√{2x-4}-√{2}}

=\lim↙{x→3}\;\;{(√{3x}-3)(√{3}x+3)(√{2x-4}+√{2}}/{(√{2x-4}-√{2})(√{2x-4}+√{2})(√{3x}+3)}

=\lim↙{x→3}\;\;{(3x-9)(√{2x-4}+√{2})}/{(2x-4-2)(√{3x+3})}

=\lim↙{x→3}\;\;{3(x-3)(√{2x-4}+√{2})}/{2(x-3)(√{3x}+3)}

=\;\;{3}/{2}×\;\;{2√{2}}/{6}=\;\;{1}/{√{2}}

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