Get the solution to "Evaluate the following limit. limx→0sin⁡ax−tan−1⁡axbx3"

facas9

Answered question

2021-01-13

Evaluate the following limit.

lim_{x \to 0} \frac{\sin a x - \tan^{- 1} a x}{b x^{3}}

Adnaan Franks

Skilled2021-01-14Added 92 answers

The given expression is

lim_{x \to 0} \frac{\sin a x - \tan^{- 1} a x}{b x^{3}}

Using Taylor series the expansion of sin x and

\tan^{- 1} x

\sin x = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{x^{2 k + 1}}{(2 k + 1)!}

\tan^{- 1} x = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{x^{2 k + 1}}{2 k + 1}

lim_{x \to 0} \frac{\sin a x - \tan^{- 1} a x}{b x^{3}}

= lim_{x \to 0} \frac{\sum_{k = 0}^{\infty} (- 1)^{k} \frac{(a x)^{2 k + 1}}{(2 k + 1)!} \sum_{k = 0}^{\infty} (- 1)^{k} \frac{(a x)^{2 k + 1}}{2 k + 1}}{b x^{3}}

= lim_{x \to 0} \frac{\sum_{k = 0}^{\infty} (- 1)^{k} (a x)^{2 k + 1} (\frac{1}{(2 k + 1)!} - \frac{1}{2 k + 1})}{b x^{3}}

= lim_{x \to 0} [\frac{(- 1)^{0} \cdot (a)^{2 \cdot 0 + 1} (x)^{2 \cdot 0 - 2} (\frac{1}{(2 \cdot 0 + 1)!} - \frac{1}{2 \cdot 0 + 1})}{b} + \frac{(- 1)^{1} \cdot (a)^{2 \cdot 1 + 1} (x)^{2 \cdot 1 - 2} (\frac{1}{(2 \cdot 1 + 1)!} - \frac{1}{2 \cdot 1 + 1}}{b} + \frac{(- 1)^{2} \cdot (a)^{2 \cdot 2 + 1} (x)^{2 \cdot 2 - 2} (\frac{1}{(2 \cdot 2 + 1)!} - \frac{1}{2 \cdot 2 + 1}}{b} + . . .]

= lim_{x \to 0} [\frac{a x^{- 2} (\frac{1}{1!} - \frac{1}{1})}{b} - \frac{a^{3} x^{0} (\frac{1}{3!} - \frac{1}{3})}{b} + \frac{a^{5} x^{2} (\frac{1}{5!} - \frac{1}{5})}{b} + . . .]

lim_{x \to 0} [0 - \frac{a^{3} (- \frac{1}{6})}{b} + \frac{a^{5} x^{2} (\frac{1}{5!} - \frac{1}{5})}{b} + . . .]

lim_{x \to 0} [0 + \frac{a^{3}}{6 b} + 0]

= \frac{a^{3}}{6 b}

Jeffrey Jordon

Expert2022-04-01Added 2605 answers

Answer is given below (on video)

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