lim-x-0-0-x-0-u-2-tan-1-1-t-dt-dt-x-x-cos-x-

Question Number 168966 by cortano1 last updated on 22/Apr/22

\lim_{x \to 0} \frac{\int_{0}^{x} (\int_{0}^{u^{2}} \tan^{- 1} (1 + t) d t) d t}{x - x \cos x} = ?

Answered by bobhans last updated on 22/Apr/22

\lim_{x \to 0} \frac{\int_{0}^{x} (\int_{0}^{u^{2}} \tan^{- 1} (1 + t) d t) d t}{x - x \cos x}

= \lim_{x \to 0} \frac{\int_{0}^{x^{2}} \tan^{- 1} (1 + t) d t}{1 - (\cos x - x \sin x)} = \lim_{x \to 0} \frac{\int_{0}^{x^{2}} \tan^{- 1} (1 + t) d t}{1 + x \sin x - \cos x}

= \lim_{x \to 0} \frac{2 x \tan^{- 1} (1 + x^{2})}{\sin x + x \cos x + \sin x}

= \frac{π}{4} . \lim_{x \to 0} \frac{2 x}{2 \sin x + x \cos x} = \frac{π}{2} \lim_{x \to 0} \frac{1}{\frac{2 \sin x}{x} + \cos x}

= \frac{π}{2} . \frac{1}{3} = \frac{π}{6}

Answered by qaz last updated on 22/Apr/22

\arctan (1 + t) = \arctan 1 + \arctan \frac{t}{2 + t} = \frac{π}{4} + \arctan (1 - \frac{1}{1 + \frac{t}{2}})

= \frac{π}{4} + \arctan [1 - (1 - \frac{t}{2} + \dots)] = \frac{π}{4} + o (1) \dots \dots . . (t \to 0)

\Rightarrow \lim_{x \to 0} \frac{\int_{0}^{x} (\int_{0}^{u^{2}} \tan^{- 1} (1 + t) d t) d t}{x - x \cos x}

= \lim_{x \to 0} \frac{\int_{0}^{x} \int_{0}^{u^{2}} [\frac{π}{4} + o (1)] dtdu}{\frac{1}{2} x^{3} + o (x^{3})}

= \lim_{x \to 0} \frac{\int_{0}^{x} [\frac{π}{4} u^{2} + o (u^{2})] du}{\frac{1}{2} x^{3} + o (x^{3})}

= \lim_{x \to 0} \frac{\frac{π}{12} x^{3} + o (x^{3})}{\frac{1}{2} x^{3} + o (x^{3})}

= \frac{π}{6}

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