Chapter 2.2, Problem 23E

To determine

To find: The value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{\cos \left(\frac{1}{x}\right)}{1+\frac{1}{x}}$ .

Answer to Problem 23E

The value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{\cos \left(\frac{1}{x}\right)}{1+\frac{1}{x}}$ are both equal to $1$ .

Explanation of Solution

Given information:The given function is $y=\frac{\cos \left(\frac{1}{x}\right)}{1+\frac{1}{x}}$ .

Calculation:

The value of $\frac{1}{x}$ approaches to $0$ from right side as the value of $x$ approaches to $\infty$ .

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

  $\begin{array}{c}\underset{x\to \infty }{\lim }y=\underset{x\to \infty }{\lim }\frac{\cos \left(\frac{1}{x}\right)}{1+\frac{1}{x}}\\ =\underset{x\to 0^{+}}{\lim }\frac{\cos \left(x\right)}{1+x}\\ =\frac{\cos 0}{1+0}\\ =1\end{array}$

The value of $\frac{1}{x}$ approaches to $0$ from left side as the value of $x$ approaches to $-\infty$ .

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

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

Therefore,the value of $\underset{x\to \infty }{\lim }y$ and $\underset{x\to -\infty }{\lim }y$ for $y=\frac{\cos \left(\frac{1}{x}\right)}{1+\frac{1}{x}}$ are both equal to $1$ .