Chapter 2.2, Problem 4QQ

(a)

To determine

To find: The domain and range of the function $f\left(x\right)=\frac{\cos x}{x}$.

Answer to Problem 4QQ

The domain and range of the function $f\left(x\right)=\frac{\cos x}{x}$ are $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ and $\left(-\infty ,\infty \right)$ respectively.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{\cos x}{x}$.

Calculation:

The function $f\left(x\right)=\frac{\cos x}{x}$ is defined for all the real values of $x$ except $x=0$.

So, the domain of the function $f\left(x\right)=\frac{\cos x}{x}$ is set of real numbers except $x=0$ or the interval $\left(-\infty ,0\right)\cup \left(0,\infty \right)$.

The value of function $f\left(x\right)=\frac{\cos x}{x}$ has no minimum and maximum for all the real values of $x$.

So, the range of the function $f\left(x\right)=\frac{\cos x}{x}$ is set of real numbers or the interval $\left(-\infty ,\infty \right)$.

(b)

To determine

To check: Whether the function $f\left(x\right)=\frac{\cos x}{x}$ is even or odd or neither.

Answer to Problem 4QQ

The function $f\left(x\right)=\frac{\cos x}{x}$ is an odd function.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{\cos x}{x}$.

Calculation:

Check the value of $f\left(-x\right)$ for the nature of $f\left(x\right)=\frac{\cos x}{x}$.

Substitute $-x$ for $x$ in the given function.

  $\begin{array}{c}f\left(x\right)=\frac{\cos x}{x}\\ f\left(-x\right)=\frac{\cos \left(-x\right)}{-x}\end{array}$

The trigonometric function cosine is even function. So, $\cos \left(-x\right)=\cos x$.

Substitute $\cos x$ for $\cos \left(-x\right)$ in the above equation.

  $\begin{array}{c}f\left(-x\right)=\frac{\cos \left(-x\right)}{-x}\\ =\frac{\cos x}{-x}\\ =-\frac{\cos \left(x\right)}{x}\\ =-f\left(x\right)\end{array}$

The function $f\left(x\right)$ holds the property $f\left(-x\right)=-f\left(x\right)$ . So, the given function is an odd function.

Therefore, the function $f\left(x\right)=\frac{\cos x}{x}$ is an odd function.

(c)

To determine

To find: The value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$.

Answer to Problem 4QQ

The value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$ is $0$.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{\cos x}{x}$.

Calculation:

Evaluate the limit by substitution.

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

Therefore, the value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$ is $0$.

(d)

To determine

To find: The value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$ with the help of Sandwich Theorem.

Answer to Problem 4QQ

The value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$ is $0$.

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{\cos x}{x}$.

Calculation:

The value of the trigonometric function cosine lies between and equal to $-1$ to $1$ i.e. $-1\le \cos x\le 1$.

Divide each side of inequality by $x$.

  $\begin{array}{c}-1\le \cos x\le 1\\ -\frac{1}{x}\le \frac{\cos x}{x}\le \frac{1}{x}\end{array}$

Apply limit to all the sides of inequality.

  $\begin{array}{l}-\frac{1}{x}\le \frac{\cos x}{x}\le \frac{1}{x}\\ \underset{x\to \infty }{\lim }\left(-\frac{1}{x}\right)\le \underset{x\to \infty }{\lim }\frac{\cos x}{x}\le \underset{x\to \infty }{\lim }\frac{1}{x}\\ 0\le \underset{x\to \infty }{\lim }\frac{\cos x}{x}\le 0\end{array}$

So, the limit of the function $f\left(x\right)=\frac{\cos x}{x}$ is $0$.

Therefore, the value of limit $\underset{x\to \infty }{\lim }f\left(x\right)$ for $f\left(x\right)=\frac{\cos x}{x}$ is $0$.