Chapter 2.2, Problem 26E

To determine

To find: The value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{x\sin x+2\sin x}{2x^2}$ .

Answer to Problem 26E

The value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{x\sin x+2\sin x}{2x^2}$ are both equal to $0$ .

Explanation of Solution

Given information:The given function is $y=\frac{x\sin x+2\sin x}{2x^2}$ .

Calculation:

Simplify the given function.

  $\begin{array}{c}y=\frac{x\sin x+2\sin x}{2x^2}\\ =\frac{x\sin x}{2x^2}+\frac{2\sin x}{2x^2}\\ =\frac{1}{2}\times \frac{\sin x}{x}+\frac{1}{x}\times \frac{\sin x}{x}\end{array}$

The right side limit of the function can be calculated as:

  $\begin{array}{c}\underset{x\to \infty }{\lim }y=\underset{x\to \infty }{\lim }\left(\frac{1}{2}\times \frac{\sin x}{x}+\frac{1}{x}\times \frac{\sin x}{x}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{1}{2}\times \frac{\sin x}{x}\right)\times \underset{x\to \infty }{\lim }\left(\frac{1}{x}\times \frac{\sin x}{x}\right)\\ =\frac{1}{2}\underset{x\to \infty }{\lim }\left(\frac{\sin x}{x}\right)+\underset{x\to \infty }{\lim }\left(\frac{1}{x}\right)\times \underset{x\to \infty }{\lim }\left(\frac{\sin x}{x}\right)\end{array}$

The value of $\underset{x\to \infty }{\lim }\frac{\sin x}{x}$ is always $0$ and the value of $\underset{x\to \infty }{\lim }\left(\frac{1}{x}\right)=0$ . So,

  $\begin{array}{c}\underset{x\to \infty }{\lim }y=\frac{1}{2}\times 0+0\times 0\\ =0\end{array}$

The left side limit of the function can be calculated as:

  $\begin{array}{c}\underset{x\to -\infty }{\lim }y=\underset{x\to -\infty }{\lim }\left(\frac{1}{2}\times \frac{\sin x}{x}+\frac{1}{x}\times \frac{\sin x}{x}\right)\\ =\underset{x\to -\infty }{\lim }\left(\frac{1}{2}\times \frac{\sin x}{x}\right)\times \underset{x\to -\infty }{\lim }\left(\frac{1}{x}\times \frac{\sin x}{x}\right)\\ =\frac{1}{2}\underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)+\underset{x\to -\infty }{\lim }\left(\frac{1}{x}\right)\times \underset{x\to -\infty }{\lim }\left(\frac{\sin x}{x}\right)\end{array}$

The value of $\underset{x\to -\infty }{\lim }\frac{\sin x}{x}$ is always $0$ and the value of $\underset{x\to -\infty }{\lim }\left(\frac{1}{x}\right)=0$ . So,

  $\begin{array}{c}\underset{x\to -\infty }{\lim }y=\frac{1}{2}\times 0+0\times 0\\ =0\end{array}$

Therefore,the value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{x\sin x+2\sin x}{2x^2}$ are both equal to $0$ .