find-the-value-of-a-for-which-the-limit-lim-x-0-sin-ax-tan-1-x-x-x-3-x-4-is-finite-and-then-evaluate-the-limit-

Question Number 194250 by horsebrand11 last updated on 01/Jul/23

find the value of a for which the limit

\lim_{x \to 0} \frac{\sin (ax) - \tan^{- 1} (x) - x}{x^{3} + x^{4}} is finite

and then evaluate the limit

Answered by qaz last updated on 01/Jul/23

\sin (a x) = a x - \frac{1}{6} a^{3} x^{3} + \dots, \arctan x = x - \frac{1}{3} x^{3} + \dots

\underset{x \to 0}{l i m} \frac{(a - 2) x + (\frac{1}{3} - \frac{1}{6} a^{3}) x^{3}}{x^{3}} = - 1, a = 2

Answered by gatocomcirrose last updated on 02/Jul/23

\lim_{x \to 0} \frac{acos (ax) - \frac{1}{1 + x^{2}} - 1}{{3 x}^{2} + {4 x}^{3}} = \frac{a - 2}{0} \to \pm \infty

a = 2 \Rightarrow \lim_{x \to 0} \frac{- a^{2} sen (ax) + \frac{2 x}{{(1 + x^{2})}^{2}}}{6 x + {12 x}^{2}}

= \lim_{x \to 0} \frac{- a^{3} \cos (ax) + \frac{2 {(1 + x^{2})}^{2} - {8 x}^{2} (1 + x^{2})}{{(1 + x^{2})}^{4}}}{6 + 24 x}

= \frac{- a^{3} + 2}{6} = \frac{- 8 + 2}{6} = - 1