NIMCET 2015 Question Paper with solutions

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Question : 20 of 120

Marks: +1, -0

The value of

\lim↙{x→a}

{√{a+2x}-√{3x}}/{√{3a+x}-2√x}

is :

$2/3$
$2/√3$
${3√3}/2$
$2/{3√3}$

Solution:

We have,

\lim↙{x→a}{√{a+2x}-√{3x}}/{√{3a+x}-2√x}

⇒\lim↙{x→a}{(√{a+2x}-√{3x})(√{a+2x}+3√{x})(√{3a+x}+2√{x})}/{(√{3a+x}+√{2x})(√{a+2x}+√{3x})+(√{3a+x}+2√{x})}

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

⇒\lim_↙{x→a}\;\;{(a-x)(√{3a+x}+2√{x})}/{3(a-x)(√{a+2x}+√{3x})}

⇒\;\;{1}/{3}×\;\;{4}/{2√{3}}=\;\;{2}/{3√{3}}

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