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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 64 of 160

Marks: +1, -0

\lim\limits_{x \to 0^{+}} \frac{x \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)}{\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \tan^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)} =

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\frac{1}{6}$

Solution:

\lim\limits_{x \to 0^{+}} \frac{x \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)}{\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \tan^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)}

∵ We know that

x \to 0^{+}

2\tan^{-1} x = \sin^{-1} \frac{2x}{1+x^{2}} = \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

3\tan^{-1} x = \tan^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)

∴ Limit becomes

\lim\limits_{x \to 0^{+}} \frac{x \cdot (2\tan^{-1} x)}{(2\tan^{-1} x)(3\tan^{-1} x)}

}

\Rightarrow \lim\limits_{x \to 0^{+}} \frac{x}{3\tan^{-1} x}

Using L Hospital rule

\lim\limits_{x \to 0^{+}} = \frac{1}{3 \times \frac{1}{1+x^{2}}} = \frac{1}{3 \times \left(\frac{1}{1+0}\right)} = \frac{1}{3}

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