Unit 1 · Topic 1.13 Beta

Removing a Discontinuity

A removable discontinuity is removed by defining f(c) to be the limit at c .

Practice 1: Implementing Mathematical ProcessesPractice 3: Justification

Part 1 · Hook

Why this matters

A bridge crew can fill a single pothole with one shovel of asphalt, but no shovel fills a canyon. Some discontinuities are potholes: one point, set to the right height, repairs them. Others, jumps and asymptotes, cannot be fixed that way. This lesson is about choosing the right value, or the right constant, to make a function whole.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $f$ is continuous at $c$ when

$f(c)$ exists, the limit at $c$ exists, and they are equal
the graph looks smooth
$f(c)$ exists
the one-sided limits exist

Show the answer

All three conditions are needed.

Correct: $f(c)$ exists, the limit at $c$ exists, and they are equal:
the graph looks smooth:
$f(c)$ exists:
the one-sided limits exist:

2. $\lim_{x \to 0} \frac{\sin(2x)}{x}$ is

2
1
0
$\frac{1}{2}$

Show the answer

$\frac{\sin(2x)}{x}$ $= 2 \cdot \frac{\sin(2x)}{2x} \to 2$ .

Correct: 2:
1:
0:
$\frac{1}{2}$ :

3. At a removable discontinuity, the limit

exists
does not exist
is unbounded
equals the function value

Show the answer

Removable: the limit exists but the value is missing or different.

Correct: exists:
does not exist:
is unbounded:
equals the function value:

Part 4 · See it

See it first

Two graphs of f(x) = kx + 1 for x < 2 and f(x) = x² − 1 for x ≥ 2. Left, with k = 2, the line reaches an open circle at (2, 5) while the parabola starts at a closed dot at (2, 3): a jump. Right, with k = 1, the line reaches (2, 3) exactly where the parabola starts, so the pieces meet and f is continuous at 2. The condition is 2k + 1 = 3, so k = 1. — Choosing $k$ so that the two pieces meet. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f$ has a removable discontinuity at $c$ , with $\lim_{x \to c} f(x) = L$ defining (or redefining) $f(c) = L$ makes $f$ continuous at $c$
A piecewise function has a constant $k$ in one or both piecescontinuity at the break point gives an equation: the left-hand limit equals the right-hand limit equals the value
There are two unknown constants and two break pointseach break point gives one equation; solve them together
The discontinuity is a jump or infiniteno single value can fix it, because the two-sided limit does not exist

Part 6 · Key ideas

Key ideas

To remove a removable discontinuity, find $L = \lim_{x \to c} f(x)$ and define $f(c) = L$ .
At a break point of a piecewise function, set the left-hand limit, the right-hand limit and the value equal. Each break point gives one equation.
A constant can only move its own piece. If the value at the point and the other piece already disagree, no constant fixes it.
A jump or an infinite discontinuity cannot be removed: the limit does not exist, so no value can equal it.
Write the result as a check: substitute your constant back and show both sides give the same number.

Part 7 · Misconception

A common mistake

The wrong idea: To make $f$ continuous at $c$ , just define $f(c)$ to be any number, since the problem is that $f(c)$ is undefined.

What actually happens: The value must equal the limit: $f(c)$ $= \lim_{x \to c} f(x)$ . And this works only when that limit exists. For a jump or an asymptote, no value of $f(c)$ makes the function continuous.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A piecewise function with a constant $k$

The function f, where k is a constant
Where	Rule for $f(x)$
$x < 2$	$kx^2 - 3$
$x \ge 2$	$2x + k$

1. Find the value of $k$ that makes $f$ continuous at $x = 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Continuity at 2 needs $\lim_{x \to 2^-} f(x)$ $= \lim_{x \to 2^+} f(x)$ $= f(2)$ . The left rule gives $4k - 3$ ; the right rule (which also gives $f(2)$ ) gives $4 + k$ . So $4k - 3 = 4 + k$ , $3k = 7$ , $k = \frac{7}{3}$ .

Answer: $\frac{7}{3}$ ≈ 2.333

2. Why is the equation $4k - 3 = 4 + k$ the right condition for continuity at $x = 2$ ?

It sets the left-hand limit equal to the right-hand limit and $f(2)$
It makes $f(2)$ defined, which is the condition that was failing at $x = 2$
It makes both rules give 0 at $x = 2$ , so $f$ has a zero there and no break
It sets the two rules equal at each $x$ , so that the pieces become one rule

Show the answer

$4k - 3$ is $\lim_{x \to 2^-} f(x)$ from the left rule. The right rule gives both $\lim_{x \to 2^+} f(x)$ and $f(2)$ , namely $4 + k$ . Setting them equal makes the limit exist and equal $f(2)$ .

Correct: It sets the left-hand limit equal to the right-hand limit and $f(2)$ : One equation covers all three conditions, because the value at 2 comes from the right rule.
It makes $f(2)$ defined, which is the condition that was failing at $x = 2$ : $f(2) = 2(2) + k$ is defined for every $k$ .
It makes both rules give 0 at $x = 2$ , so $f$ has a zero there and no break: Nothing needs to be 0; the two sides just need to match.
It sets the two rules equal at each $x$ , so that the pieces become one rule: The rules are compared only at $x = 2$ , where the pieces meet.

3. If $k = 1$ , what happens at $x = 2$ ?

A removable discontinuity, a hole at height 1
A jump; the sides approach 1 and 5
An infinite discontinuity at $x = 2$
Nothing; $f$ is continuous at $x = 2$

Show the answer

With $k = 1$ : $\lim_{x \to 2^-} f(x)$ $= 1 \cdot 4 - 3$ $= 1$ and $\lim_{x \to 2^+} f(x)$ $= 4 + 1$ $= 5$ . Finite one-sided limits that differ make a jump.

A removable discontinuity, a hole at height 1: A removable discontinuity needs the two sides to agree; here they are 1 and 5.
Correct: A jump; the sides approach 1 and 5: Left rule $x^2 - 3 \to 1$ , right rule $2x + 1 \to 5$ : they differ.
An infinite discontinuity at $x = 2$ : Both rules are polynomials, so nothing is unbounded.
Nothing; $f$ is continuous at $x = 2$ : Only $k = \frac{7}{3}$ makes the pieces meet.

Table

Four functions, each undefined at one point

Four functions, each with a removable discontinuity at the point shown
Function	Formula	Undefined at
$a$	$a(x)$ $= \frac{x^2 - 5x + 6}{x - 2}$	$x = 2$
$b$	$b(x) = \frac{\sqrt{x + 9} - 3}{x}$	$x = 0$
$c$	$c(x) = \frac{\sin(4x)}{x}$	$x = 0$
$d$	$d(x)$ $= \frac{e^{2x} - 1}{e^x - 1}$	$x = 0$

4. What value should $b(0)$ be given so that $b$ is continuous at $x = 0$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Multiply by the conjugate: $\frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)}$ $= \frac{1}{\sqrt{x + 9} + 3}$ for $x \ne 0$ . The limit is $\frac{1}{6}$ , so define $b(0) = \frac{1}{6}$ .

Answer: $\frac{1}{6}$ ≈ 0.167

5. Which discontinuity can be removed by defining or changing a single value of the function?

$\frac{|x - 1|}{x - 1}$ at $x = 1$
$\frac{x + 1}{x - 1}$ at $x = 1$
$\frac{x^2 - 1}{x - 1}$ at $x = 1$
$\frac{1}{(x - 1)^2}$ at $x = 1$

Show the answer

A single value can fix a discontinuity only when the limit exists. $\frac{x^2 - 1}{x - 1} = x + 1$ near 1, so its limit is 2; defining the value 2 there removes the break.

$\frac{|x - 1|}{x - 1}$ at $x = 1$ : The one-sided limits are $-1$ and $1$ : a jump. No single value makes them agree.
$\frac{x + 1}{x - 1}$ at $x = 1$ : It is unbounded near 1; no value can fix that.
Correct: $\frac{x^2 - 1}{x - 1}$ at $x = 1$ : The limit exists (2), so the hole can be filled.
$\frac{1}{(x - 1)^2}$ at $x = 1$ : It is unbounded near 1; no value can fix that.

6. Let $f(x) = x^2 + k$ for $x < 3$ , $f(3) = 5$ , and $f(x) = 2x + 1$ for $x > 3$ . Which value of $k$ makes $f$ continuous at $x = 3$ ?

$k = -2$ , so the left side approaches 7, like the right side
No value of $k$ ; the right side approaches 7, but $f(3) = 5$
$k = -4$ , so the left side approaches $f(3) = 5$
$k = 2$ , the difference between 7 and $f(3) = 5$

Show the answer

Continuity needs the left limit, the right limit and $f(3)$ all equal. The right limit is $2(3) + 1 = 7$ , which differs from $f(3) = 5$ whatever $k$ is. So no $k$ works.

$k = -2$ , so the left side approaches 7, like the right side: The limit would then be 7, but $f(3) = 5$ , so the third condition fails.
Correct: No value of $k$ ; the right side approaches 7, but $f(3) = 5$ : $k$ only moves the left piece; the mismatch between 7 and 5 stays.
$k = -4$ , so the left side approaches $f(3) = 5$ : The left side would approach 5, but the right side approaches 7: a jump.
$k = 2$ , the difference between 7 and $f(3) = 5$ : Subtracting values does not fix a mismatch between the right side and $f(3)$ .

Part 9 · Summary

Summary

A removable discontinuity is removed by defining

f(c)

to be the limit at

c

. A piecewise function is made continuous at a break point by choosing constants so the two one-sided limits and the value agree; each break point gives one equation. Jumps and infinite discontinuities cannot be removed by choosing a value.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

A piecewise function with a constant kk

Four functions, each undefined at one point

Summary

What comes next

Connections

A piecewise function with a constant $k$