Explained: "Evaluate the following limits using Taylor series. limx→0tan−1⁡x−xx3"

aflacatn

Answered question

2021-02-04

Evaluate the following limits using Taylor series.

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

cheekabooy

Skilled2021-02-05Added 83 answers

Given,
We have to evaluate the limit of the given expression by using the Taylor series.

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

Formula Used
The Taylor series of

\tan^{- 1} x

\tan^{- 1} x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .

Calculation:

lim_{x \to 0} \frac{\tan^{- 1} x - x}{x^{3}} = lim_{x \to 0} \frac{(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .) - x}{x^{3}}

= lim_{x \to 0} \frac{x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . . - x}{x^{3}}

= lim_{x \to 0} \frac{- \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .}{x^{3}}

= lim_{x \to 0} \frac{x^{3} (- \frac{1}{3!} + \frac{x^{2}}{5!} - \frac{x^{4}}{7!} + . . .)}{x^{3}}

= lim_{x \to 0} (- \frac{1}{3!} + \frac{x^{2}}{5!} - \frac{x^{4}}{7!} + . . .)

= - \frac{1}{3!} + 0 - 0 + . . .

= - \frac{1}{6}

Hence the limit of the given expression using the Taylor's series is

- \frac{1}{6}

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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