Chapter 8, Problem 49RQ

To determine

To Find: The value of the limit $\underset{x\to 0}{\lim }\frac{\sin x-x}{x^3}$ .

Answer to Problem 49RQ

The value of the limit is $\frac{1}{6}$ .

Explanation of Solution

Calculation:

Consider the given limit is,

  $\underset{x\to 0}{\lim }\frac{\sin x-x}{x^3}$

Consider the series is,

  $\begin{array}{l}\sin x={\displaystyle \sum _{n=0}^{\infty }\frac{\left(-1\right)x^{2n+1}}{\left(2n+1\right)!}}\\ \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{...}\\ \sin x-x=-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{...}\end{array}$

Then, the limit is obtained as,

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{\sin x-x}{x^3}=\frac{-1}{6}+\underset{x\to 0}{\lim }\left[\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{...}\right]\\ =-\frac{1}{6}+0\\ =\frac{1}{6}\end{array}$