Chapter 3.3, Problem 50E

(a)

To determine

To sketch: The graph of $f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}$ and describe the discontinuity at 0.

Explanation of Solution

Definition used:

If the left hand limit of $f\left(x\right)$ and the right hand limit of $f\left(x\right)$ exists but are not equal, then the function $f\left(x\right)$ said to be jump discontinuity. That is, $\underset{x\to a^{+}}{\lim }f\left(x\right)\ne \underset{x\to a^{-}}{\lim }f\left(x\right)$.

Use online graphing calculator to draw the graph of the function $f\left(x\right)$ as shown in Figure 1.

The function $f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}$ is jump discontinuity at 0.

Use online graphing calculator to draw the graph of the function $f\left(x\right)$ as shown in Figure 1.

From Figure 1, it is observed that the left hand limit of $f\left(x\right)$ and the right hand limit of $f\left(x\right)$ exists but are not equal.

Use definition stated above, it can be concluded that the function $f\left(x\right)$ is jump discontinuity at 0.

Therefore, the function $f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}$ is jump discontinuity at 0.

(b)

To determine

To calculate: The left and right limits of $f\left(x\right)$ at 0

Answer to Problem 50E

The left limit of $f\left(x\right)$ at 0 is $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-\frac{1}{\sqrt{2}}$ and the right limit of $f\left(x\right)$ at 0 is $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\frac{1}{\sqrt{2}}$.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}$.

Result Used:

The value of $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$.

Calculation:

The given can be expressed as follows,

$\begin{array}{c}f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}\\ =\frac{x}{\sqrt{2{\sin }^{2}x}}\left[Q1-\cos 2x=2{\sin }^{2}x\right]\\ =\frac{x}{\sqrt{2}\sqrt{{\sin }^{2}x}}\\ =\frac{x}{\sqrt{2}\left|\sin x\right|}\left[Q\sqrt{a^2}=\left|a\right|\right]\end{array}$

The left limit of $f\left(x\right)$ is computed as follows,

$\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{x}{\sqrt{2}\left(-\sin x\right)}\\ =\underset{x\to 0^{-}}{\lim }\frac{-x}{\sqrt{2}\left(\sin x\right)}\\ =\underset{x\to 0^{-}}{\lim }\frac{1}{\sqrt{2}\left(\frac{\sin x}{x}\right)}\\ =\frac{-1}{\sqrt{2}\underset{x\to 0^{-}}{\lim }\left(\frac{\sin x}{x}\right)}\end{array}$

Use Result stated above and obtain the left limit,

$\begin{array}{c}\underset{x\to 0^{-}}{\lim }f\left(x\right)=\frac{-1}{\sqrt{2}\left(1\right)}\\ =\frac{-1}{\sqrt{2}}\end{array}$

Therefore, the left limit is $\underset{x\to 0^{-}}{\lim }f\left(x\right)=-\frac{1}{\sqrt{2}}$.

The right limit of $f\left(x\right)$ is computed as follow,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{x}{\sqrt{2}\left(+\sin x\right)}\\ =\underset{x\to 0^{+}}{\lim }\frac{1}{\sqrt{2}\left(\frac{\sin x}{x}\right)}\\ =\frac{1}{\sqrt{2}\underset{x\to 0^{+}}{\lim }\left(\frac{\sin x}{x}\right)}\end{array}$

Use Result stated above and obtain the left limit,

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }f\left(x\right)=\frac{1}{\sqrt{2}\left(1\right)}\\ =\frac{1}{\sqrt{2}}\end{array}$

Therefore, the right limit is $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\frac{1}{\sqrt{2}}$.

Note that, the left limit of $f\left(x\right)$ and the right limit of $f\left(x\right)$ exits but are not equal.

That is, $\underset{x\to 0^{+}}{\lim }f\left(x\right)\ne \underset{x\to 0^{-}}{\lim }f\left(x\right)$.

Use definition stated above, it is conclude that the function $f\left(x\right)$ is jump discontinuity at 0.

Therefore, the function $f\left(x\right)=\frac{x}{\sqrt{1-\cos 2x}}$ is jump discontinuity at 0.