Continuity

A function is continuous at a point when its limit there is its value there. Most marks in this topic come from applying that definition to a piecewise function with an unknown constant, from reading off where a formula is continuous, and from telling a removable discontinuity, which a single value would repair, from a jump or an infinite one. Continuity on a closed interval also pays off in the Intermediate Value Theorem: a continuous function that changes sign between $a$ and $b$ has a zero between them, which shows an equation has a solution without solving it.

Start practising Browse all 28 questions

In this topic

Taught: Before the midterm
Questions: 28
Total marks: 69
Suggested time: 117 min
Difficulty: Introductory to Capstone
Show the 6 hardest

Key results

Continuity at a point: $f \text{ is continuous at } a \iff \lim_{x \to a} f(x) = f(a)$
Both sides must exist: $f(a)$ must be defined and the limit must be a real number.
Continuity at a join: $f \text{ is continuous at } a \iff \lim_{x \to a^{-}} f(x) = f(a) = \lim_{x \to a^{+}} f(x)$
Where formulas are continuous: $\lim_{x \to a} f(x) = f(a) \ \text{ for every } a \text{ in the domain of } f$
For $f$ built from polynomials, roots, exponentials, logarithms and trigonometric functions by $+$ , $-$ , $\times$ , $\div$ and composition, with a one-sided limit at an endpoint of the domain.
Kinds of discontinuity: $\begin{aligned} &\text{removable: } \lim_{x \to a} f(x) \text{ exists} \\ &\text{jump: } \lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x) \\ &\text{infinite: } \lim_{x \to a^{-}} f(x) \text{ or } \lim_{x \to a^{+}} f(x) \text{ is } \pm\infty \end{aligned}$
For a function that is not continuous at $x = a$ : removable when $f(a)$ is undefined or differs from the limit, a jump when both one-sided limits are real numbers.
The Intermediate Value Theorem: $Y \text{ between } f(a) \text{ and } f(b) \implies f(c) = Y \text{ for some } c \in [a, b]$
When $a < b$ and $f$ is continuous on the closed interval $[a, b]$ ; in particular, if $f(a)$ and $f(b)$ have opposite signs, $f$ has a zero strictly between $a$ and $b$ .

Read the theoryUBC Math 100 §2.3 Limits and continuityCLP-1 §1.6 Continuity

Worked example

Standard2 marks · 4 min

Find the value of the constant $c$ that makes $f$ continuous everywhere.

f(x) = \begin{cases} cx^{2} - 1, & x < 2 \\[4pt] 3x + c, & x \geq 2 \end{cases}

Strategy

Polynomials are continuous, so only the join matters. Continuity at the join is one equation: the two pieces must agree there.

Working

$\lim_{x \to 2^{-}} f(x) = c(2)^{2} - 1 = 4c - 1$
From the left, the first piece applies.
$\lim_{x \to 2^{+}} f(x) = f(2) = 3(2) + c = 6 + c$
From the right, the second piece applies, and it also gives f(2).
$4c - 1 = 6 + c \quad\Longrightarrow\quad 3c = 7 \quad\Longrightarrow\quad c = \tfrac{7}{3}$
Continuity at 2 forces the two to agree.

Answer

c = \frac{7}{3}

Check

With c = 7/3 both pieces meet at height 25/3 ≈ 8.33 when x = 2: 28/3 − 1 = 6 + 7/3.

Where this goes wrong

Substituting x = 2 into only one piece, or setting the pieces equal as functions rather than at the single point x = 2. The condition is about the values at the join only.

On the paper

At a join, write the three conditions: $f(a)$ is defined, both one-sided limits agree, and they equal $f(a)$ . Before using the Intermediate Value Theorem, say why $f$ is continuous on the closed interval, then name the theorem.

Leans on

One-sided limits of a piecewise function, from Limits

Your turn

Introductory2 marks · 3 min

Find the value of the constant $c$ that makes $f$ continuous at $x = 0$ .

f(x) = \begin{cases} \dfrac{\sin(6x)}{2x}, & x < 0 \\[10pt] 2x + c, & x \geq 0 \end{cases}

Answer this question

Practice questions

5 of the 28, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
A function $f$ is continuous at $x = 2$ , and $\lim_{x \to 2} f(x) = 5$ . Which of the following must be true?
Strategy
Continuity at a point is one equation: the limit equals the value. Given the limit and given continuity, the value is forced.
Working
$\text{continuous at } 2 \iff \lim_{x \to 2} f(x) = f(2)$
The definition.
$f(2) = \lim_{x \to 2} f(x) = 5$
Substitute the given limit.
Answer
$f(2) = 5$
The answer is f of 2 equals 5.
Check
f(x) = 5 + |x − 2| is continuous with limit 5 at 2 and has f(2) = 5, yet it is not 5 near 2 and has no f′(2).
Where this goes wrong
Choosing “f is constant near 2”. A limit of 5 says the values get close to 5, not that they equal 5. Continuity adds nothing about differentiability either.
Leans on
The definition of a limit, from Limits
Try a similar question
Find the value of $a$ that makes $f$ continuous at $x = 1$ .
Answer this question
Introductory2 marks · 3 min
On which set is the function below continuous?
$f(x) = \log\bigl(4 - x^{2}\bigr) + \frac{1}{x - 1}$
Strategy
Logarithms and rational functions are continuous wherever they are defined, so the sum is continuous exactly where both terms are defined.
Working
$4 - x^{2} > 0 \iff x^{2} < 4 \iff -2 < x < 2$
A logarithm needs a strictly positive input; 0 is not allowed.
$x - 1 \neq 0 \iff x \neq 1$
A denominator cannot be zero.
$(-2, 1) \cup (1, 2)$
Keep the points that pass both tests: the interval (−2, 2) with x = 1 removed.
Answer
$(-2, 1) \cup (1, 2)$
The answer is every x strictly between −2 and 2 except x = 1.
Check
f(1.5) = log 1.75 + 2 ≈ 2.56 is defined, while f(2.5) would need log(−2.25) and f(1) would divide by 0.
Where this goes wrong
Including the endpoints ±2, or forgetting the gap at 1. log 0 is undefined, so the logarithm needs 4 − x² strictly positive, and the second term removes x = 1 from the middle of that interval.
Leans on
Domain of a logarithm, from Functions, graphs and inverses
Try a similar question
On which set is $g(x) = \dfrac{1}{\sqrt{x^{2} - 4}}$ continuous?
Answer this question
Standard2 marks · 3 min
Find the value of $a$ that makes $f$ continuous at $x = 1$ .
$f(x) = \begin{cases} 2x + 1, & x < 1 \\[4pt] a x^{2} - 3, & x \geq 1 \end{cases}$
Strategy
A piecewise function is continuous at a join point when the two pieces approach the same value there. Evaluate both one-sided limits and force them to agree.
Working
$\lim_{x \to 1^{-}} f(x) = 2(1) + 1 = 3$
Approaching from the left uses the first piece.
$\lim_{x \to 1^{+}} f(x) = a(1)^{2} - 3 = a - 3$
Approaching from the right uses the second piece, which also defines f(1).
$a - 3 = 3$
Continuity requires the two sides to meet.
$a = 6$
Solve. With a = 6 both sides give 3, and f(1) = 3 as well.
Answer
$a = 6$
The answer is a = 6.
Check
Either side of 1, at x = 0.999 and 1.001, the pieces give 2.998 and about 3.012, both close to 3.
Where this goes wrong
Setting a(1)² = 3 and answering a = 3, which drops the −3 from the second piece. Substitute into the whole expression, not just the term containing a.
Leans on
One-sided limits, from Limits
Try a similar question
The function below is continuous everywhere. Find $a + b$ .
Answer this question
Challenging3 marks · 6 min
A tax rule charges 15% of income on incomes up to $\$50\,000$ . On larger incomes it charges 25% of the whole income, minus a fixed deduction of $D$ dollars. $D$ is chosen so that the tax $T(x)$ on an income of $x$ dollars is a continuous function: crossing the threshold never makes the tax jump. Find $D$ , and then enter the tax owed on an income of $\$80\,000$ .
$T(x) = \begin{cases} 0.15x, & 0 \leq x \leq 50\,000 \\[4pt] 0.25x - D, & x > 50\,000 \end{cases}$
Strategy
Each rule is a line, so the only place the tax could jump is the threshold. Continuity there is one equation, which fixes D; then evaluate the rule that applies to $80,000.
Working
$T(50\,000) = 0.15(50\,000) = 7500$
At the threshold itself the first rule applies.
$\lim_{x \to 50\,000^{+}} T(x) = 0.25(50\,000) - D = 12\,500 - D$
Just above the threshold the second rule applies.
$12\,500 - D = 7500 \quad\Longrightarrow\quad D = 5000$
Continuity: the limit from above must equal the value at the threshold.
$T(80\,000) = 0.25(80\,000) - 5000 = 20\,000 - 5000 = 15\,000$
$80,000 is above the threshold, so use the second rule.
$0.25x - 5000 = 7500 + 0.25(x - 50\,000)$
A check that also explains the rule: 15% on the first $50,000 and 25% only on the part above it, 7500 + 7500 = 15,000.
Answer
$D = 5000, \qquad T(80\,000) = \$15\,000$
The answer is D is 5000 dollars, and the tax on 80,000 dollars is 15,000 dollars.
Check
$15,000 is 18.75% of $80,000, between the two rates of 15% and 25%, as an average rate must be.
Where this goes wrong
Stopping at D = 5000, or charging 25% on the whole income and forgetting to subtract D. Finding D is the continuity step; the question then asks for the tax that rule gives at $80,000.
Try a similar question
Find the value of the constant $c$ that makes $f$ continuous everywhere.
Answer this question
Capstone5 marks · 10 min
Find the values of the constants $a$ and $b$ that make $f$ continuous everywhere. Justify your answer using the definition of continuity.
$f(x) = \begin{cases} \dfrac{x^{3} + 1}{x + 1}, & x < -1 \\[10pt] ax + b, & -1 \leq x \leq 2 \\[6pt] \dfrac{x^{2} + 5x - 14}{x - 2}, & x > 2 \end{cases}$
Write your reasoning in full, then compare it against the model solution.
Strategy
Away from the joins every piece is continuous, so continuity everywhere comes down to two conditions, one at each join. Find the outer one-sided limits by factoring, set each equal to the value of the middle piece, and solve the pair of equations.
Working
On x < −1 the denominator x + 1 is never zero, on x > 2 the denominator x − 2 is never zero, and the middle piece is a polynomial. So f is continuous everywhere except possibly at x = −1 and x = 2.
$\lim_{x \to -1^{-}} \frac{x^{3} + 1}{x + 1} = \lim_{x \to -1^{-}} \bigl(x^{2} - x + 1\bigr) = 1 + 1 + 1 = 3$
Sum of cubes: x³ + 1 = (x + 1)(x² − x + 1). Cancel, then substitute.
$\lim_{x \to 2^{+}} \frac{x^{2} + 5x - 14}{x - 2} = \lim_{x \to 2^{+}} (x + 7) = 9$
Factor as (x + 7)(x − 2) and cancel x − 2.
$x = -1:\ -a + b = 3, \qquad x = 2:\ 2a + b = 9$
At each join the outer one-sided limit must equal the value of the middle piece, which is also its limit from the other side.
$3a = 6 \quad\Longrightarrow\quad a = 2, \quad b = 5$
Subtract the first equation from the second, then substitute back.
$\lim_{x \to -1} f(x) = 3 = f(-1), \qquad \lim_{x \to 2} f(x) = 9 = f(2)$
With a = 2 and b = 5 both joins meet the definition: the limit exists and equals the value. The middle piece is the line through (−1, 3) and (2, 9).
Answer
$a = 2, \qquad b = 5$
The answer is a equals 2 and b equals 5.
What earns the marks
Explains that each piece is continuous on its own open interval (neither outer denominator is zero there), so only x = −1 and x = 2 need checking.
Finds the left-hand limit at −1 by factoring x³ + 1 = (x + 1)(x² − x + 1) and cancelling: the limit is 3.
Finds the right-hand limit at 2 by factoring x² + 5x − 14 = (x + 7)(x − 2) and cancelling: the limit is 9.
Uses the middle piece for the values f(−1) = −a + b and f(2) = 2a + b, and sets each equal to the matching one-sided limit.
Solves −a + b = 3 and 2a + b = 9 to get a = 2 and b = 5, and concludes that at each join the limit exists and equals the value.
Check
At x = −1.01 and x = 2.01 the outer pieces give 3.0301 and 9.01, close to 2x + 5 at the joins, 3 and 9.
Where this goes wrong
Substituting x = −1 into (x³ + 1)/(x + 1), getting 0/0 and treating the left side as 0. The outer pieces never reach their joins; what matters is their one-sided limits, found after factoring.
Leans on
Limits by factoring and cancelling, from Limits
Try a similar question
The function below is continuous everywhere. Find $a + b$ .
Answer this question

Browse all 28 questions

Needs
Limits
Limit laws, the indeterminate form 0/0, one-sided and infinite limits, limits at infinity and asymptotes.
30 questions
Leads to
The definition of the derivative
Secants and tangents, average and instantaneous velocity, the difference quotient and differentiability.
27 questions
Leads to
The Mean Value Theorem
Rolle’s theorem, the MVT, counting roots and proving inequalities.
11 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

What earns the marks

Check

Where this goes wrong

Leans on

Try a similar question

Related topics

Limits

The definition of the derivative

The Mean Value Theorem