Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q8: Discontinuities of $\frac{|x|}{x}$ for $x\ne0,$ with $f(0)=0$

Exercise 5.1Points of discontinuity · Section 5.2

The question

Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} \dfrac{|x|}{x}, & \text{if } x \ne 0 \\[2mm] 0, & \text{if } x = 0 \end{cases}$

Exam answer

For $x < 0$ ⁠, $f(x) = \frac{-x}{x} = -1$ ⁠, and for $x > 0$ ⁠, $f(x) = \frac{x}{x} = 1$ ⁠. Constant functions are continuous, so $f$ is continuous at every $x \ne 0$ ⁠.

At $x = 0$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(0 - h) \\ &= \lim_{h\to0}\frac{|-h|}{-h} = \lim_{h\to0}\frac{h}{-h} \\ &= -1 \\ \text{RHL} &= \lim_{h\to0} f(0 + h) \\ &= \lim_{h\to0}\frac{|h|}{h} = \lim_{h\to0}\frac{h}{h} \\ &= 1 \end{aligned}

and $f(0) = 0$ ⁠. Since $\text{LHL} \ne \text{RHL}$ ⁠, $f$ is not continuous at $x = 0$ ⁠.

Hence $x = 0$ is the only point of discontinuity of $f$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

$\frac{|x|}{x}$ only records the sign of $x$ ⁠: it is $1$ for every positive $x$ and $-1$ for every negative $x.$ So $f$ is constant on each side of 0, hence continuous away from 0. At 0 the values just to the left are all $-1$ and just to the right are all $1$ ⁠; the one-sided limits differ, so $f$ jumps at 0, whatever its value there.

Given, to find, and the plan

Given: $f(x) = \frac{|x|}{x}$ for $x \ne 0$ and $f(0) = 0.$
To find: Every point of the domain at which $f$ is not continuous.
Plan: (1) Domain; simplify $\frac{|x|}{x}$ for $x > 0$ and $x < 0.$ (2) Points $c \ne 0.$ (3) The point $c = 0.$ (4) Collect the answer.

Basics you need

1Continuity at a point; point of discontinuity

$f$ is continuous at $c$ (in its domain) when the left-hand limit, the right-hand limit and $f(c)$ are all equal; otherwise $c$ is a point of discontinuity (Definition 1, Section 5.2). Lesson: continuity at a point

2Absolute value

$|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0.$ Basics: piecewise functions

3Constant functions are continuous; limits are local

A constant function is continuous everywhere (Example 5). If $f$ is constant on an open interval around $c,$ then $\lim_{x\to c} f(x)$ is that constant. Basics: what a limit is

4One-sided limits with a small step $h$

With $h > 0$ ⁠: LHL at $c$ is $\lim_{h\to0} f(c-h)$ and RHL is $\lim_{h\to0} f(c+h).$ Basics: left-hand and right-hand limits

Solution

Domain, and what $\frac{|x|}{x}$ really is

The first formula covers every $x \ne 0$ (where dividing by $x$ is allowed) and the second covers $x = 0,$ so $f$ is defined on all of $\R.$ Now simplify:
- If $x > 0$ ⁠: $|x| = x,$ so $\dfrac{|x|}{x} = \dfrac{x}{x} = 1.$
- If $x < 0$ ⁠: $|x| = -x,$ so $\dfrac{|x|}{x} = \dfrac{-x}{x} = -1.$
So
$f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$
The fraction looks complicated but takes only two values. Cancelling $x$ is legitimate because $x \ne 0$ on this piece. The second line of the definition is needed precisely because $\frac{|x|}{x}$ has no meaning at 0 (it would be $\frac00$ ⁠).

$|x| = x$ for $x \ge 0,$ $-x$ for $x < 0.$ Basics: piecewise functions
Points $c \ne 0$ ⁠: continuous

If $c > 0$ ⁠: every $x$ with $|x - c| < c$ is positive, so $f(x) = 1$ on that interval. Hence $\lim_{x\to c} f(x) = 1 = f(c).$

If $c < 0$ ⁠: every $x$ with $|x - c| < -c$ is negative, so $f(x) = -1$ there, and $\lim_{x\to c} f(x) = -1 = f(c).$

So $f$ is continuous at every $c \ne 0.$

Around a non-zero point, a small enough interval never reaches 0, so $f$ is constant on it. A constant function is continuous (Example 5), and the limit only sees that interval.
The point $c = 0$

Value. $f(0) = 0$ (given directly).

Left-hand limit. For $h > 0,$ $-h < 0,$ so
$\lim_{x\to0^-} f(x) = \lim_{h\to0} \frac{\lvert -h\rvert }{-h} = \lim_{h\to0} \frac{h}{-h} = -1.$
Right-hand limit. For $h > 0,$
$\lim_{x\to0^+} f(x) = \lim_{h\to0} \frac{|h|}{h} = \lim_{h\to0} \frac{h}{h} = 1.$
The one-sided limits $-1$ and $1$ differ, so $\lim_{x\to0} f(x)$ does not exist and $f$ is not continuous at $x = 0$ .

The graph of $f$ ⁠: the line $y = -1$ for $x < 0$ (open circle at $(0, -1)$ ⁠), the line $y = 1$ for $x > 0$ (open circle at $(0, 1)$ ⁠), and the single filled point $(0, 0).$ The graph jumps by 2 at $x = 0.$

Here $f(0) = 0$ matches neither side, but that is not even the main problem: since the two one-sided limits differ, no value of $f(0)$ could make $f$ continuous at 0. This is the classic jump of the sign function.

Basics: left-hand and right-hand limits
Collect the answer

$f$ is continuous at every $c \ne 0$ (Step 2) and not at $c = 0$ (Step 3). So $x = 0$ is the only point of discontinuity.

Steps 2 and 3 together cover every real number.

Result

The only point of discontinuity is $x = 0$ : the left-hand limit there is $-1,$ the right-hand limit is $1$ and $f(0) = 0.$ At every other real number $f$ is continuous. (Textbook answer: "Discontinuous at $x = 0$ ".)

Check it with numbers

$f(-0.001) = \frac{0.001}{-0.001} = -1,$ $f(0) = 0,$ $f(0.001) = \frac{0.001}{0.001} = 1.$ However small the step, the values on the two sides stay 2 apart. Away from 0, for example at $c = 2$ ⁠: $f(1.999) = f(2) = f(2.001) = 1.$

Common mistakes

" $\frac{|x|}{x} = 1$ always." For negative $x,$ $|x| = -x,$ so the fraction is $-1.$
Thinking the discontinuity could be removed by choosing $f(0)$ better. The one-sided limits differ, so no value at 0 works.
Writing $\lvert -h\rvert = -h.$ For $h > 0,$ $\lvert -h\rvert = h.$

What to take away

Simplify a formula on each side of the critical point before testing: $\frac{|x|}{x}$ is just the sign of $x.$
Locally constant means continuous: around any non-zero point $f$ does not change.
When the one-sided limits differ, the discontinuity is a jump and cannot be fixed by redefining the single value $f(c).$

Every fact used, and where it comes from

Fact	Used in	Source
Continuity at $c$ ⁠; point of discontinuity	Steps 2–4	Definition 1, Section 5.2 (Lesson: continuity at a point)
$\|x\| = x$ for $x \ge 0,$ $-x$ for $x < 0$	Steps 1, 3	Definition of modulus (Basics: piecewise functions)
Constant functions are continuous	Step 2	Example 5
A limit depends only on values near the point	Step 2	Class XI Limits (Basics: what a limit is)
One-sided limits as $\lim_{h\to0} f(c\mp h),$ $h > 0$	Step 3	Class XI Limits (Basics: one-sided limits)
Values in the check box and the graph coordinates	Check; figure	Recomputed in the site's verification script

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Exercise 5.1 · Q8: Discontinuities of ∣x∣x\frac{|x|}{x}x∣x∣​ for x≠0,x\ne0,x=0, with f(0)=0f(0)=0f(0)=0

Basics you need

Solution

Domain, and what ∣x∣x\frac{|x|}{x}x∣x∣​ really is

Points c≠0c \ne 0c=0⁠: continuous

The point c=0c = 0c=0

Collect the answer

Every fact used, and where it comes from

Exercise 5.1 · Q8: Discontinuities of $\frac{|x|}{x}$ for $x\ne0,$ with $f(0)=0$

Domain, and what $\frac{|x|}{x}$ really is

Points $c \ne 0$ ⁠: continuous

The point $c = 0$