calculate-lim-x-0-arctan-1-x-3-pi-4-xsin-x-2-

Question Number 66696 by mathmax by abdo last updated on 18/Aug/19

c a l c u l a t e {l i m}_{x \to 0} \frac{a r c t a n (1 + x^{3}) - \frac{π}{4}}{x s i n (x^{2})}

Commented by kaivan.ahmadi last updated on 18/Aug/19

{l i m}_{x \to 0} \frac{3 x^{2} / (1 + {(1 + x^{3})}^{2})}{3 x^{2}} = {l i m}_{x \to 0} (1 / 1 + {(1 + x^{3})}^{2}) = 1 / 2

Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19

l e t n a m e d i t L

L = \lim_{x \to 0} \frac{a r c t a n (1 + x^{3}) - \frac{π}{4}}{x^{3}} . \frac{x^{2}}{s i n (x^{2})}

N o w \lim_{x \to 0} \frac{x^{2}}{s i n (x^{2})} = \lim_{u \to 0} \frac{1}{\frac{s i n u}{u}} = 1 (u = x^{2})

\lim_{x \to 0} \frac{a r c t a n (1 + x^{3}) - \frac{π}{4}}{x^{3}} = \lim_{v \to 0} \frac{a r c t a n (1 + v) - \frac{π}{4}}{v} = \lim_{v \to 0} \frac{1}{1 + {(1 + v)}^{2}} = \frac{1}{2} (v = x^{3})

S o L = \frac{1}{2}

Commented by mathmax by abdo last updated on 19/Aug/19

l e t f (u) = a r c t a n (1 + u) w e h a v e

f (u) = f (0) + u f^{^{'}} (0) + o (u^{2}) b u t f (0) = \frac{π}{4}

f^{^{'}} (u) = \frac{1}{1 + {(1 + u)}^{2}} \Rightarrow f^{^{'}} (0) = \frac{1}{2} \Rightarrow f (u) = \frac{π}{4} + \frac{1}{2} u + o (u^{2}) \Rightarrow

a r c t a n (1 + x^{3}) = \frac{π}{4} + \frac{x^{3}}{2} + o (x^{6}) \Rightarrow a r c t a n (1 + x^{3}) - \frac{π}{4} \sim \frac{x^{3}}{2} (x \to 0)

s l s o x s i n (x^{2}) \sim x^{3} \Rightarrow {l i m}_{x \to 0} \frac{a r c t a n (1 + x^{3}) - \frac{π}{4}}{x s i n (x^{2})} = {l i m}_{x \to 0} \frac{x^{3}}{2 x^{3}}

= \frac{1}{2} .

Answered by Cmr 237 last updated on 18/Aug/19

Answered by Cmr 237 last updated on 18/Aug/19

w i t h t h e l i m i t e d d e v e l p m e n t w e

h a v e :

a r c t a n (1 + x^{3}) = \frac{π}{4} + \frac{x^{3}}{2} + o (x^{3})

s i n (x^{2}) = x^{2} + o (x^{2})

s o

l i m \underset{x \to 0}{} \frac{a r c t a n (1 + x^{3}) - \frac{π}{4}}{x s i n (x^{2})} = l i \underset{}{m} \underset{x \to 0}{} \frac{\frac{π}{4} + \frac{x^{3}}{2} - \frac{π}{4}}{x \times x^{2}}

= \frac{1}{2} .

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