Chapter 8.7, Problem 53E

To determine

To Find: The limit of $\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}$

Answer to Problem 53E

The value of the limit is $\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}=\frac{1}{120}$

Explanation of Solution

Given information:

  $\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}$

Concept Used:

Maclaurin series for $\sin x$

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{..}$

Calculation:

The limit of $\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}$ is obtained as:

UsingMaclaurin series for $\sin x$

  $\begin{array}{c}\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}=\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{6}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{..}-x+\frac{1}{6}{x}^3}{x^5}\\ =\underset{x\to 0}{\lim }\frac{\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{..}}{x^5}\\ =\underset{x\to 0}{\lim }\frac{\frac{1}{5!}-\frac{x^2}{7!}\mathrm{..}}{1}\\ \underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}=\frac{1}{120}\end{array}$

The value of the limit is $\underset{x\to 0}{\lim }\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}=\frac{1}{120}$