Chapter 5 · Continuity and Differentiability

Example 4: $x^3+3$ with $f(0)=1$ is not continuous at 0

ExamplesContinuity at a point · Section 5.2

The question

Show that the function $f$ given by

f(x) = \begin{cases} x^3 + 3, & \text{if } x \ne 0 \\ 1, & \text{if } x = 0 \end{cases}

is not continuous at $x = 0$ ⁠.

Exam answer

f(0) = 1

For $x \ne 0$ ⁠, $f(x) = x^3 + 3$ ⁠.

\begin{aligned} \text{LHL} &= \lim_{x\to 0^-}(x^3 + 3) \\ &= 0 + 3 = 3 \end{aligned}

\begin{aligned} \text{RHL} &= \lim_{x\to 0^+}(x^3 + 3) \\ &= 0 + 3 = 3 \end{aligned}

So $\lim_{x\to 0} f(x) = 3 \ne 1 = f(0)$ ⁠.

$\therefore\ f$ is not continuous at $x = 0$ ⁠. Hence proved.

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

The one idea this problem is built on

A limit at a point is decided by the values near the point, never by the value at it. Near 0 this function is $x^3 + 3$ ⁠, so the graph heads for height 3 from both sides. But the function has been given the value 1 at 0. The limit exists, the value exists, and they disagree, so continuity fails. The graph has a hole at $(0, 3)$ and a stray point at $(0, 1)$ ⁠.

Given, to show, and the plan

Given: $f(x) = x^3 + 3$ for every $x \ne 0$ ⁠, and $f(0) = 1$ ⁠.
To show: $f$ is not continuous at $x = 0$ ⁠. (We also justify the textbook's remark that 0 is the only point of discontinuity.)
Plan: Find $f(0)$ from the second line. Find the left-hand and right-hand limits at 0 from the first line. Show the common limit is not $f(0)$ ⁠. Then show $f$ is continuous at every $c \ne 0$ ⁠.

Basics you need

1A limit ignores the value at the point

$\lim_{x\to c} f(x)$ depends only on $f(x)$ for $x$ near $c$ with $x \ne c$ ⁠. Changing $f(c)$ alone never changes the limit. Basics: what a limit is

2Reading a piecewise definition

Each line applies only where its condition holds. " $x \ne 0$ " covers every real number except 0; " $x = 0$ " covers only 0. Basics: piecewise functions

3One-sided limits and polynomials

The limit at 0 exists when the left-hand and right-hand limits exist and agree. For a polynomial $p$ ⁠, $\lim_{x\to c} p(x) = p(c)$ from either side (algebra of limits, Class XI). Basics: left-hand and right-hand limits

4Definition 1: continuity at a point

For $c$ in the domain, $f$ is continuous at $c$ iff $\lim_{x\to c} f(x) = f(c)$ ⁠. If not, $f$ is discontinuous at $c$ and $c$ is a point of discontinuity. Lesson: continuity at a point

Solution

Is 0 in the domain? Find $f(0)$

The two lines together cover every real number (every $x$ is either 0 or not 0), so the domain is $\R$ and 0 is in it. At $x = 0$ only the second line applies:
$f(0) = 1.$
The first line says "if $x \ne 0$ ", so it may not be used at 0 at all. Putting 0 into $x^3 + 3$ would give 3, which is not the value of this function at 0.

In a piecewise definition, the value at a point comes only from the line whose condition that point satisfies. Basics: piecewise functions
Left-hand limit at 0

For $x < 0$ we have $x \ne 0$ ⁠, so $f(x) = x^3 + 3$ ⁠. Hence
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-}(x^3 + 3) = 0^3 + 3 = 3.$
The left-hand limit uses only numbers just below 0, all of which satisfy " $x \ne 0$ ". So the value $f(0) = 1$ never enters this calculation (Basic 1). Since $x^3 + 3$ is a polynomial, its limit is found by putting $x = 0$ into that formula.

$\lim x^3 = (\lim x)^3$ by the product rule, and $\lim 3 = 3$ ⁠. Basics: one-sided limits

Right-hand limit at 0

For $x > 0$ also $x \ne 0$ ⁠, so again $f(x) = x^3 + 3$ ⁠:

\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+}(x^3 + 3) = 0^3 + 3 = 3.

Both one-sided limits are 3, so $\lim_{x\to 0} f(x) = 3$ ⁠. Some values near 0:

$x$	$f(x)$	which line
$-0.1$	$2.999$	$x \ne 0$
$-0.01$	$2.999999$	$x \ne 0$
$0$	$1$	$x = 0$
$0.01$	$3.000001$	$x \ne 0$

The table makes the trouble visible: the values on both sides crowd around 3, but the value at 0 itself is 1, far away from them.

Compare: the limit is not the value
$\begin{aligned} \lim_{x\to 0^-} f(x) &= 3, \\ \lim_{x\to 0^+} f(x) &= 3, \\ f(0) &= 1. \end{aligned}$
The limit exists and equals 3, but $3 \ne 1 = f(0)$ ⁠. Definition 1 fails, so $f$ is not continuous at $x = 0$ .

For continuity all three numbers must agree. Here the two limits agree with each other but not with the value, which is the second kind of failure shown in Fig 5.2 of the textbook (limits equal, value different).

Lesson: continuity at a point
Why 0 is the only point of discontinuity

Take any $c \ne 0$ ⁠. Every $x$ with $|x - c| < \frac{|c|}{2}$ has the same sign as $c$ ⁠, so $x \ne 0$ and $f(x) = x^3 + 3$ there. Near $c$ ⁠, on both sides, $f$ is just the polynomial $x^3 + 3$ ⁠, so
$\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = c^3 + 3.$
Also $f(c) = c^3 + 3$ ⁠, because $c \ne 0$ ⁠. The three numbers agree, so $f$ is continuous at every $c \ne 0$ ⁠.

The textbook states this in one sentence; here is the reason. A limit at $c$ only looks at $x$ close to $c$ (Basic 1), and if we stay within half the distance from $c$ to 0, we never reach 0, the one place where the special value 1 is used.

If $|x - c| < \frac{|c|}{2}$ ⁠, then $x$ lies strictly between $\frac{c}{2}$ and $\frac{3c}{2}$ ⁠, two numbers with the same sign as $c$ ⁠. For $c = -0.2$ ⁠, every $x$ in $(-0.3, -0.1)$ is negative.

The graph of

f

(horizontal unit 92 px, vertical unit 44 px). The curve

y = x^3 + 3

has a hole at

(0, 3)

⁠: the graph heads there from both sides but the point is not on it. The function's actual value at 0 is the rose dot

(0, 1)

⁠.

Result

$\lim_{x\to0} f(x) = 3$ but $f(0) = 1$ ⁠, so $f$ is not continuous at $x = 0$ ; it is continuous at every other real number, so $x = 0$ is its only point of discontinuity. This matches the textbook's solution.

Check it with numbers

$f(-0.01) = 2.999999$ and $f(0.01) = 3.000001$ ⁠: just $0.000001$ from 3, but about 2 away from the value $f(0) = 1$ ⁠. No choice of $x$ close to 0 brings $f(x)$ close to 1.

Common mistakes

Computing $f(0)$ from the wrong line. $0^3 + 3 = 3$ is not $f(0)$ ⁠; the line " $x \ne 0$ " does not apply at 0.
Letting $f(0)$ affect the limit. The limit is 3 no matter what value is assigned at 0.
Saying "the limit does not exist". It does exist (both sides give 3). What fails is the equality of the limit with the value.

What to take away

Continuity can fail even when the limit exists: the value must also equal it.
This discontinuity is removable: redefining $f(0) = 3$ would make $f$ continuous at 0 (it would then just be $x^3 + 3$ ⁠).
To show a point $c$ of a piecewise function is "safe", find a small interval around $c$ on which only one formula is used.

Every fact used, and where it comes from

Fact	Used in	Source
Each line of a piecewise rule applies only where its condition holds	Steps 1–3, 5	Definition of a piecewise function (Basics: piecewise functions)
Limit of a polynomial at $c$ is its value at $c$	Steps 2, 3, 5	Algebra of limits, Class XI (Limits and Derivatives)
Limit depends only on values near $c$ ⁠, $x \ne c$	Steps 2, 5	Meaning of a limit (Basics: what a limit is)
Continuous at $c$ iff $\lim_{x\to c} f(x) = f(c)$ ⁠; point of discontinuity	Steps 4, 5	Definition 1, Section 5.2 (Lesson: continuity at a point)
Limits equal but value different (Fig 5.2)	Step 4	Section 5.2, second introductory function

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Example 4: x3+3x^3+3x3+3 with f(0)=1f(0)=1f(0)=1 is not continuous at 0

Basics you need

Solution

Is 0 in the domain? Find f(0)f(0)f(0)

Left-hand limit at 0

Right-hand limit at 0

Compare: the limit is not the value

Why 0 is the only point of discontinuity

Every fact used, and where it comes from

Example 4: $x^3+3$ with $f(0)=1$ is not continuous at 0

Is 0 in the domain? Find $f(0)$