Question

Evaluate the following limits using Taylor series. lim_{xto0}frac{tan^{-1}x-x}{x^3}

Series
Evaluate the following limits using Taylor series.
$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

2021-02-05
Given,
We have to evaluate the limit of the given expression by using the Taylor series.
$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$
Formula Used
The Taylor series of $$\tan^{-1}x$$ is $$\tan^{-1}x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
Calculation:
$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}=\lim_{x\to0}\frac{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...)-x}{x^3}$$
$$=\lim_{x\to0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...-x}{x^3}$$
$$=\lim_{x\to0}\frac{-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}{x^3}$$
$$=\lim_{x\to0}\frac{x^3(-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+...)}{x^3}$$
$$=\lim_{x\to0}(-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+...)$$
$$=-\frac{1}{3!}+0-0+...$$
$$=-\frac{1}{6}$$
Hence the limit of the given expression using the Taylor's series is $$-\frac{1}{6}$$