Question

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

Series
ANSWERED
asked 2021-02-04
Evaluate the following limits using Taylor series.
\(\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}\)

Answers (1)

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 \(-\frac{1}{6}\)
0
 
Best answer

expert advice

Need a better answer?
...