limx→∞(x+1)tan−1x+2x+3−cot−1x+3x+1=

limx→∞(x+1)tan−1x+2x+3−cot−1x+3x+1=limx→∞(x+1)tan−1x+2x+3−tan−1x+1x+3=limx→∞(x+1)tan−1x+2x+3−x+1x+31+x+2x+3x+1x+3=limx→∞(x+1)tan−1x+3(x+3)2+(x+2)(x+1)=limx→∞(x+1)tan−1x+32x2+9x+11=limx→∞(x+1)tan−1x+32x2+9x+11x+32x2+9x+11x+32x2+9x+11=limx→∞(x+1)x+32x2+9x+11=12

limx→∞(x+1)tan−1x+2x+3−cot−1x+3x+1=

limx→∞(x+1)tan−1x+2x+3−cot−1x+3x+1=limx→∞(x+1)tan−1x+2x+3−tan−1x+1x+3=limx→∞(x+1)tan−1x+2x+3−x+1x+31+x+2x+3x+1x+3=limx→∞(x+1)tan−1x+3(x+3)2+(x+2)(x+1)=limx→∞(x+1)tan−1x+32x2+9x+11=limx→∞(x+1)tan−1x+32x2+9x+11x+32x2+9x+11x+32x2+9x+11=limx→∞(x+1)x+32x2+9x+11=12

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$\lim_{x \to \infty} (x + 1) \{\tan^{- 1} (\frac{x + 2}{x + 3}) - \cot^{- 1} (\frac{x + 3}{x + 1})\} =$

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$\frac{- 1}{2}$

$\frac{1}{2}$

-1

answer is C.

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Detailed Solution

$\begin{matrix} \lim_{x \to \infty} (x + 1) \{\tan^{- 1} (\frac{x + 2}{x + 3}) - \cot^{- 1} (\frac{x + 3}{x + 1})\} = \lim_{x \to \infty} (x + 1) \{\tan^{- 1} (\frac{x + 2}{x + 3}) - \tan^{- 1} (\frac{x + 1}{x + 3})\} \\ = \lim_{x \to \infty} (x + 1) \tan^{- 1} (\frac{\frac{x + 2}{x + 3} - \frac{x + 1}{x + 3}}{1 + \frac{x + 2}{x + 3} \frac{x + 1}{x + 3}}) \\ = \lim_{x \to \infty} (x + 1) \tan^{- 1} (\frac{x + 3}{{(x + 3)}^{2} + (x + 2) (x + 1)}) \\ = \lim_{x \to \infty} (x + 1) \tan^{- 1} (\frac{x + 3}{2 x^{2} + 9 x + 11}) \\ = \lim_{x \to \infty} (x + 1) \frac{\tan^{- 1} (\frac{x + 3}{2 x^{2} + 9 x + 11})}{(\frac{x + 3}{2 x^{2} + 9 x + 11})} (\frac{x + 3}{2 x^{2} + 9 x + 11}) \\ = \lim_{x \to \infty} (x + 1) (\frac{x + 3}{2 x^{2} + 9 x + 11}) \\ = \frac{1}{2} \end{matrix}$