Chapter 1.2, Problem 11E

To determine

To find: The value of limit.

Answer to Problem 11E

The limit is 0.

Explanation of Solution

Given information: The limit is $\underset{x\to -\infty }{\lim }\frac{\sin x}{x}$

Calculation:

The given expression is:

  $\underset{x\to -\infty }{\lim }\frac{\sin x}{x}$

The range of $\text{sinx=(-1,1)}$

  $-1\le \sin x\le 1$

Divide by $\text{x}$ . Because we are finding the limit as $\text{x}$ approaches negative $x<0$ , remember that:

  $-\frac{1}{x}\ge \frac{\sin x}{x}\ge \frac{1}{x}$

The inequality indications must be reversed when dividing by a negative integer.

By used this sandwich theorem, recast with less than signs for the sake of sanity

  $\frac{1}{x}\le \frac{\sin x}{x}\le -\frac{1}{x}$

Identify the "slices of bread boundaries’’.

  $\underset{x\to -\infty }{\lim }\frac{1}{x}=0$

  $\underset{x\to -\infty }{\lim }\frac{1}{x}=0$ is provided at the section's commencement.

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\left(-\frac{1}{x}\right)=-\underset{x\to -\infty }{\lim }\frac{1}{x}\\ -1\cdot 0\\ =0\end{array}$

The sandwich theorem holds true because the limits of the two "breads" are equal, and the limit of the "meat" is also equal.

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{1}{x}=\underset{x\to -\infty }{\lim }\left(-\frac{1}{x}\right)\\ =0\\ \Rightarrow \underset{x\to -\infty }{\lim }\frac{\sin x}{x}=0\end{array}$

Therefore the required limit is 0.