JEE MainMathematicsModerateLimits

Find the absolute value
of $\lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) - 2\tan^{- 1}x}$

Show answer and solution

\lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{\pi - 2\tan^{- 1}x - 2\tan^{- 1}x} = \lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{4\left( \frac{\pi}{4} - \tan^{- 1}x \right)}

Apply LHR

= \lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4}{4\left( \frac{- 1}{1 + x^{2}} \right)} = - 2

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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