Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q9: Points of discontinuity: $\frac{x}{|x|}$ for $x < 0$ ⁠, $-1$ for $x \ge 0$

Exercise 5.1Continuity of piecewise functions · 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 < 0 \\[2mm] -1, & \text{if } x \ge 0 \end{cases}

Exam answer

For $x < 0$ ⁠, $|x| = -x$ ⁠, so $f(x) = \frac{x}{-x} = -1$ ⁠. For $x > 0$ ⁠, $f(x) = -1$ ⁠. 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|} = -1 \\ \text{RHL} &= \lim_{h\to0} f(0 + h) \\ &= \lim_{h\to0}(-1) = -1 \end{aligned}

and $f(0) = -1$ ⁠. So $\text{LHL} = \text{RHL} = f(0)$ ⁠: $f$ is continuous at $x = 0$ ⁠.

Hence $f$ is continuous at every real number: there is no point of discontinuity.

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

The one idea this problem is built on

Simplify before you test. For a negative $x$ ⁠, the modulus is $|x| = -x$ ⁠, so $\dfrac{x}{|x|} = \dfrac{x}{-x} = -1$ ⁠. The first formula is just $-1$ in disguise, and the second formula is $-1$ too. So $f(x) = -1$ for every real $x$ ⁠: $f$ is a constant function, and a constant function has no breaks. The formula $\frac{x}{|x|}$ is undefined at $0$ ⁠, but that does not matter, because at $x = 0$ the question uses the second formula, which gives $f(0) = -1$ ⁠.

Given, to find, and the plan

Given: A function $f$ with two pieces: $\frac{x}{|x|}$ for $x < 0$ and the constant $-1$ for $x \ge 0$ ⁠.
To find: Every point of the domain at which $f$ is not continuous.
Plan: (1) Find the domain. (2) Simplify the first piece using the definition of $|x|$ ⁠. (3) Check continuity at every point other than the junction $x = 0$ ⁠. (4) Test the junction $x = 0$ with the left-hand limit, the right-hand limit and the value $f(0)$ ⁠, each separately. (5) Conclude.

Basics you need

1Reading a piecewise function

Each formula applies only on its own part of the real line. To find $f(c)$ ⁠, first decide which condition $c$ satisfies, then use that formula only. Basics: piecewise functions

2The modulus

$|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x < 0$ ⁠. For example $|-3| = -(-3) = 3$ ⁠. So for a negative number, $|x|$ is minus that number, which is positive.

3Left-hand and right-hand limits

$\lim_{x\to c^-} f(x)$ uses only $x$ slightly less than $c$ ⁠; $\lim_{x\to c^+} f(x)$ uses only $x$ slightly greater than $c$ ⁠. The limit at $c$ exists exactly when both exist and are equal. Basics: one-sided limits

4Continuity at a point (Definition 1)

For a point $c$ in the domain, $f$ is continuous at $c$ when the left-hand limit, the right-hand limit and $f(c)$ all exist and are equal. If not, $c$ is a point of discontinuity. Basics: what a limit is

5Constant functions are continuous

If $f(x) = k$ for all $x$ ⁠, then $\lim_{x\to c} f(x) = k = f(c)$ at every $c$ (Example 5 of the chapter).

Solution

Find the domain of $f$

Every real number satisfies exactly one of the two conditions $x < 0$ and $x \ge 0$ ⁠. The second formula, $-1$ ⁠, is defined for every $x$ ⁠. The first formula, $\frac{x}{|x|}$ ⁠, needs $|x| \ne 0$ ⁠, that is $x \ne 0$ ⁠; it is only used for $x < 0$ ⁠, where this holds. So $f$ is defined at every real number: its domain is $\R$ ⁠.

Continuity is only asked about points of the domain. If $\frac{x}{|x|}$ were also used at $x = 0$ ⁠, the point $0$ would drop out of the domain. Checking the conditions shows that this does not happen: $0$ belongs to the second piece.

A fraction is defined only when its denominator is not zero. Basics: piecewise functions
Simplify the first piece

Take any $x < 0$ ⁠. By the definition of the modulus, $|x| = -x$ ⁠, and $-x \ne 0$ ⁠. So
$\frac{x}{|x|} = \frac{x}{-x} = -1 \qquad (x < 0).$
The second piece is $-1$ already. Therefore
$f(x) = -1 \quad \text{for every real } x.$
The first formula looks as if it could take many values, but its numerator and denominator have the same size and opposite signs when $x$ is negative, so the quotient is always $-1$ ⁠. Spotting this turns the whole question into a question about a constant function.

$|x| = -x$ for $x < 0$ ⁠, as the chapter writes it in Example 3: $|x| = -x$ if $x < 0$ ⁠, $|x| = x$ if $x \ge 0$ ⁠.
Points away from the junction

Let $c \ne 0$ ⁠. Whether $c$ is negative or positive, every $x$ near $c$ gives $f(x) = -1$ (Step 2), so
$\lim_{x\to c} f(x) = \lim_{x\to c} (-1) = -1 = f(c).$
So $f$ is continuous at every $c \ne 0$ ⁠.

This is Example 5 of the chapter (a constant function is continuous at every real number). We still treat the junction $x = 0$ on its own in the next step, because that is where the definition of $f$ changes formula, and where a careless reader might expect trouble from $\frac{x}{|x|}$ ⁠.

The limit of a constant is that constant. Lesson: continuity at a point
Test the junction $x = 0$

Value. $0 \ge 0$ ⁠, so the second formula applies: $f(0) = -1$ ⁠.

Left-hand limit. As $x \to 0^-$ ⁠, $x < 0$ ⁠, so the first formula applies, and by Step 2 it equals $-1$ ⁠:
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} \frac{x}{|x|} = \lim_{x\to 0^-} (-1) = -1.$
Right-hand limit. As $x \to 0^+$ ⁠, $x > 0$ ⁠, so the second formula applies:
$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} (-1) = -1.$
Compare. Left-hand limit $= -1$ ⁠, right-hand limit $= -1$ ⁠, $f(0) = -1$ ⁠. All three are equal, so $f$ is continuous at $x = 0$ .

Definition 1 asks for three things to agree at the junction. We compute each one with the formula that really applies on that side, rather than guessing from a single formula. The undefined value of $\frac{x}{|x|}$ at $0$ never enters: the left-hand limit only uses negative $x$ ⁠, and the value at $0$ comes from the other formula.

Left-hand limit: only $x < c$ ⁠; right-hand limit: only $x > c$ ⁠. Basics: one-sided limits
Conclude

$f$ is continuous at every $c \ne 0$ (Step 3) and at $c = 0$ (Step 4), that is, at every point of its domain $\R$ ⁠. So $f$ has no point of discontinuity.

"Find all points of discontinuity" can have the answer "none". The figure shows why: the graph is one horizontal line, drawn without lifting the pen.

The graph of

f

is the single line

y = -1

⁠. Blue: the part given by

\frac{x}{|x|}

for

x < 0

⁠. Gold: the part given by

-1

for

x \ge 0

⁠. The filled dot

(0, -1)

is the value

f(0)

⁠; there is no gap. One square is one unit on both axes.

Result

For $x < 0$ ⁠, $\frac{x}{|x|} = -1$ ⁠, so $f(x) = -1$ for every real $x$ ⁠. At the only junction $x = 0$ ⁠: left-hand limit $-1$ ⁠, right-hand limit $-1$ ⁠, $f(0) = -1$ ⁠. $f$ is continuous everywhere: there is no point of discontinuity. (Textbook answer: "No point of discontinuity".)

Check it with numbers

$f(-5) = \frac{-5}{|-5|}$ $= \frac{-5}{5} = -1$ ⁠, $f(-0.001) = \frac{-0.001}{0.001} = -1$ ⁠, $f(0) = -1$ ⁠, $f(0.001) = -1$ ⁠, $f(7) = -1$ ⁠. Values just left of $0$ ⁠, at $0$ and just right of $0$ are all exactly $-1$ ⁠: no jump. ✓

Common mistakes

" $\frac{x}{|x|}$ is undefined at 0, so $f$ is discontinuous at 0." The first formula is never used at $0$ ⁠; $f(0) = -1$ comes from the second formula.
Taking $|x| = x$ for negative $x$ ⁠. That gives $\frac{x}{x} = 1$ and a false jump from $1$ to $-1$ ⁠. For $x < 0$ ⁠, $|x| = -x$ ⁠.
Mixing this up with Question 8. There $f(x) = \frac{|x|}{x}$ for $x \ne 0$ and $f(0) = 0$ ⁠: the left-hand limit is $-1$ ⁠, the right-hand limit is $1$ ⁠, and $0$ is a point of discontinuity. Here both sides give $-1$ ⁠.
Skipping the junction. Even when each piece is continuous, the point where the formula changes must be tested.

What to take away

Simplify each piece first. A piece that looks complicated may be a constant.
A formula's undefined points only matter where that formula is actually used.
At a junction, compute the left-hand limit, the right-hand limit and the value separately, each with the formula that applies there.

Every fact used, and where it comes from

Fact	Used in	Source
$\|x\| = -x$ for $x < 0$ ⁠, $\|x\| = x$ for $x \ge 0$	Steps 2, 4	Definition of the modulus function; Example 3, Section 5.2
A fraction is defined only where its denominator is non-zero	Step 1	Arithmetic of real numbers
Continuity at $c$ ⁠: left-hand limit $=$ right-hand limit $= f(c)$	Steps 3, 4	Definition 1, Section 5.2 (Lesson: continuity at a point)
A constant function is continuous at every real number	Step 3	Example 5, Section 5.2
Left-hand and right-hand limits; the limit exists when they agree	Step 4	Class XI, Limits and Derivatives; recalled in Section 5.2

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Exercise 5.1 · Q9: Points of discontinuity: x∣x∣\frac{x}{|x|}∣x∣x​ for x<0x < 0x<0⁠, −1-1−1 for x≥0x \ge 0x≥0

Basics you need

Solution

Find the domain of fff

Simplify the first piece

Points away from the junction

Test the junction x=0x = 0x=0

Conclude

Every fact used, and where it comes from

Exercise 5.1 · Q9: Points of discontinuity: $\frac{x}{|x|}$ for $x < 0$ ⁠, $-1$ for $x \ge 0$

Find the domain of $f$

Test the junction $x = 0$