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UPSC NDA Math Model Paper 8

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Question : 38 of 120

Marks: +1, -0

\lim\limits_{x\to\infty}[\sqrt{4x^{2}+x-3}-2x]

is equal to

1/6
6
1/4
4

Solution:

\lim\limits_{x\to\infty}[\sqrt{4x^{2}+x-3}-2x]

this is ∞ - ∞ form

Here we rationalize numerator

\lim\limits_{x\to\infty}[\sqrt{4x^{2}+x-3}-2x]

\times \frac{\sqrt{4x^{2}+x-3}+2x}{\sqrt{4x^{2}+x-3}+2x}

\lim\limits_{x\to\infty}\frac{[\sqrt{4x^{2}+x-3}^{2}-(2x)^{2}]}{\sqrt{4x^{2}+x-3}+2x}

\lim\limits_{x\to\infty}\frac{[4x^{2}+x-3-4x^{2}]}{\sqrt{4x^{2}+x-3}+2x}

\lim\limits_{x\to\infty}\frac{x-3}{\sqrt{4x^{2}+x-3}+2x}

, this is

\frac{\infty}{\infty}

form

Here we divide the numerator and denominator by x

\lim\limits_{x\to\infty}\frac{1-\frac{3}{x}}{\sqrt{4+\frac{1}{x}-\frac{3}{x^{2}}}+2}=\frac{1}{4}

Hence, option 3 is the correct answer.

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