MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: continuity and the Intermediate Value Theorem (MATH 140)

This sheet is not a summary of section 2.5 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on continuity and the Intermediate Value Theorem in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter is short and the marking is strict: a continuity question is worth the three conditions written out, a seam question the two one-sided limits computed, and an IVT question the hypotheses checked BEFORE the theorem is named. Every value below is exact and done by hand, as on the exam.

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The thread of the chapter

Continuity is CHECKED, never assumed: $\lim_{x\to a} f(x) = f(a)$ hides three checks made in order, the type of a discontinuity is read from the one-sided LIMITS, a seam is computed with limits and never by plugging into a $\frac{0}{0}$ piece, and the IVT is invoked only after continuity on the whole CLOSED interval is written, to give existence and nothing more.

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Continuity at a point
One-sided continuity
Continuity on an interval
Removable discontinuity
Jump discontinuity
Infinite discontinuity
Piecewise-defined function
Continuity of a composition
Continuous extension
Intermediate Value Theorem
Fixed point
Sign chart

The essentials

One equation, three checks, and the limits decide the type

• $f$ is continuous at $a$ when $\lim_{x\to a} f(x) = f(a)$ , which means, in this order: (1) $f(a)$ is defined; (2) the limit exists, both one-sided limits finite and EQUAL; (3) the limit equals $f(a)$ .
• Removable: the limit exists, but $f(a)$ is undefined or different. One value, the limit, repairs it.
• Jump: both one-sided limits finite and different. No value repairs it.
• Infinite: at least one one-sided limit is $\pm\infty$ . No value repairs it.
• None of the three: $\sin\frac{1}{x}$ at $0$ oscillates between $-1$ and $1$ without a limit and without blowing up. The list is not complete.
• At an endpoint of a closed interval, only the one-sided continuity makes sense: from the right at $a$ , from the left at $b$ .

-3

the curve heads for

2

but the filled dot sits at

4

(removable); at

0

the sides arrive at

1

and

3

(jump); at

3

the curve escapes to infinity (infinite).

The type is read from the one-sided LIMITS, never from the condition that failed first: $\frac{x^2 - 9}{x - 3}$ and $\frac{1}{x - 3}$ both fail condition 1 at $3$ , and one is removable while the other is infinite.

Where to look, and what the IVT really says

• Every function built from polynomials, roots, $|x|$ , $e^x$ , $\ln x$ and trigonometric functions is continuous on its domain. Only the EXCLUDED points and the points where a formula CHANGES are candidates.
• Composition: if $\lim_{x\to a} g(x) = b$ and $f$ is continuous at $b$ , then $\lim_{x\to a} f(g(x)) = f(b)$ . The hypothesis is on the OUTER function at $b$ .
• IVT: $f$ continuous on the CLOSED interval $[a, b]$ and $N$ strictly between $f(a)$ and $f(b)$ give some $c$ in $(a, b)$ with $f(c) = N$ .
• The IVT gives AT LEAST one $c$ . Never exactly one, never where, never which values are NOT taken.
• Consequence behind every sign table: continuous and never zero on an interval implies one constant sign on that interval.

A marker gives the IVT marks for three written sentences: the function and the interval, why it is continuous on the closed interval, the two end values with their signs.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Reading a point: the one-sided limits against the value

Compute the left limit, the right limit and $f(a)$ , then read the verdict in the last column. The verdict never depends on which condition you checked first.

Left and right limits	Value f(a)	Verdict at a
both equal $L$	$f(a) = L$	continuous
Example: $x^2 + 2x + 4$ at $2$ : both sides give $12$ , and so does the value.
both equal $L$	undefined or $\ne L$	removable
Example: $\frac{x^3 - 8}{x - 2}$ at $2$ : limit $12$ , value undefined; set $f(2) = 12$ .
finite, different	anything	jump
Example: $\frac{x^2 - 1}{\|x - 1\|}$ at $1$ : $-2$ on the left, $2$ on the right.
one is $\pm\infty$	anything	infinite
Example: $\frac{x - 1}{\|x\| - 1}$ at $-1$ : numerator near $-2$ , denominator near $0$ .
no limit, bounded	anything	none of the three
Example: $\sin\frac{1}{x}$ at $0$ : equals $1$ at $x = \frac{2}{5\pi}$ and $-1$ at $x = \frac{2}{7\pi}$ , and so on closer to $0$ .

Only the second line can be repaired by choosing a value, and the only value that works is the limit $L$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Plugging the seam into a piece that gives zero over zero

the whole question

What not to write

“At $x = 1$ the left piece $\frac{x^2 - 1}{x - 1}$ gives $\frac{0}{0}$ , so $g$ cannot be continuous at $1$ .”

What to write

“For $x < 1$ , $\frac{x^2 - 1}{x - 1} = x + 1$ , so $\lim_{x\to 1^-} g(x) = 2$ . Continuity at $1$ requires $g(1) = 2$ .”

Why: Continuity compares a LIMIT with a value. The left piece is never evaluated at $1$ , which is outside its interval; only its limit matters, and a $\frac{0}{0}$ form is simplified first.

2. Calling a function continuous because its limit exists

1 to 2 marks

What not to write

“ $\lim_{x\to 2}\frac{x^2 - 4}{x - 2} = 4$ , so $f$ is continuous at $2$ .”

What to write

“The limit is $4$ , but $f(2)$ is undefined: condition 1 fails, and $f$ has a removable discontinuity at $2$ . Setting $f(2) = 4$ makes it continuous.”

Why: The existence of the limit is condition 2 only. Condition 1 (a value) and condition 3 (the value equals the limit) must be checked and written.

3. Classifying by the first condition that fails

1 mark per discontinuity

What not to write

“ $\frac{1}{x - 3}$ is undefined at $3$ , like $\frac{x^2 - 9}{x - 3}$ , so both discontinuities are removable.”

What to write

“ $\frac{x^2 - 9}{x - 3} \to 6$ , removable; $\frac{1}{x - 3} \to \infty$ as $x \to 3^+$ , infinite. The type is decided by the one-sided limits.”

Why: Removable means that a finite limit EXISTS. An undefined value says nothing about the limit, which is the only thing that can be repaired.

4. Removing an absolute value without cases

2 marks

What not to write

“ $\frac{x^2 - 1}{|x - 1|} = x + 1 \to 2$ , so the discontinuity at $1$ is removable.”

What to write

“For $x > 1$ the quotient is $x + 1 \to 2$ ; for $x < 1$ it is $-(x + 1) \to -2$ . The one-sided limits differ: jump discontinuity.”

Why: $|x - 1| = x - 1$ on one side only. An absolute value at the point studied ALWAYS means two computations, one per side.

5. Keeping one root of a quadratic condition on a parameter

1 mark

What not to write

“Continuity at $1$ gives $c^2 = 3c - 2$ ; dividing by $c$ ... so $c = 2$ .”

What to write

“ $c^2 - 3c + 2 = (c - 1)(c - 2) = 0$ , so $c = 1$ or $c = 2$ , and both values make the two sides equal: $1 = 1$ and $4 = 4$ .”

Why: When the parameter enters squared, the answer is a set. Factor, never divide by an unknown, and check each value in both pieces.

6. Applying the IVT across a point where the function is undefined

the whole question

What not to write

“ $f(x) = \frac{1}{x - 2}$ has $f(0) = -\frac{1}{2} < 0$ and $f(3) = 1 > 0$ , so $f$ has a zero in $(0, 3)$ .”

What to write

“ $f$ is undefined at $2$ , which lies in $[0, 3]$ , so $f$ is not continuous on $[0, 3]$ and the IVT does not apply. In fact $\frac{1}{x - 2}$ is never $0$ .”

The curve starts below the axis and ends above it, but it jumps at

x = 2

instead of crossing: no zero on

[0, 3]

, and no contradiction with the IVT.

Why: The sign changes by jumping through infinity, not by crossing $0$ . A sign change is evidence of a root only for a function continuous on the WHOLE closed interval.

7. Invoking the IVT without writing its hypotheses

1 to 2 method marks

What not to write

“ $h(0) < 0$ and $h(1) > 0$ , so by the IVT there is a root.”

What to write

“ $h(x) = e^x + x - 3$ is continuous on $[0, 1]$ as a sum of continuous functions; $h(0) = -2 < 0$ and $h(1) = e - 2 > 0$ since $e > 2$ . By the IVT, $h(c) = 0$ for some $c$ in $(0, 1)$ .”

Why: The continuity on the closed interval IS the hypothesis; without it the conclusion is not proved. The exact values show the signs without a calculator.

8. Reading uniqueness, or a location, into the IVT

1 mark, and the credibility of the rest of the answer

What not to write

“ $f(0) = 1$ , $f(6) = 5$ and $f$ is continuous, so $f(c) = 3$ for exactly one $c$ , and $1 \le f(x) \le 5$ on $[0, 6]$ .”

What to write

“By the IVT, $f(c) = 3$ for AT LEAST one $c$ in $(0, 6)$ . The theorem says nothing about how many such $c$ exist, nor about the values outside $[1, 5]$ .”

Why: The IVT is an existence theorem. Uniqueness needs another argument, which comes later in the course; the function $1 + \frac{2x}{3} + 1.8\sin\frac{2\pi x}{3}$ meets $y = 3$ five times.

Which method to choose

Which tool, by the VERB of the question

Read what the question asks before computing anything: the verb picks the tool

If is $f$ continuous at $a$ ? → write the three conditions in order: the value, the two one-sided limits, the equality

Example: $\frac{x^2 - 4}{x - 2}$ with $f(2) = 5$ : limit $4 \ne 5$ , condition 3 fails
If classify the discontinuity → compute BOTH one-sided limits; they alone decide

Example: $\frac{x^2 + 3x}{|x|}$ at $0$ : $-3$ and $3$ , jump
If find the parameter(s) that make f continuous → each piece continuous on its open interval, then one equation per seam, limits computed by simplifying

Example: $kx^2 + 3x$ and $2x^3 - kx$ at $1$ : $k + 3 = 2 - k$ , $k = -\frac{1}{2}$
If can f be extended, or redefined at one point? → only if the limit exists and is finite; the value is that limit

Example: $\frac{\sqrt x - 3}{x - 9}$ at $9$ : extend with $\frac{1}{6}$
If show that an equation has a solution, or that f takes a value → move everything to one side, check continuity on a closed interval, find a sign change, cite the IVT

Example: $2^x - 3x$ : $+1$ at $0$ , $-1$ at $1$ , a root in $(0, 1)$
If solve an inequality, or give the sign of f → zeros AND points of discontinuity cut the line; one test value per piece

Example: $\frac{x^2 - x - 2}{x - 3} > 0$ on $(-1, 2) \cup (3, \infty)$
If compute a limit of a composition → limit of the inner function, then continuity of the OUTER function at that limit

Example: $\arctan\frac{x^2 - 4}{4x - 8} \to \arctan 1 = \frac{\pi}{4}$ at $2$

No derivative anywhere in this chapter: monotonicity, uniqueness of a root and differentiability are later chapters, and a solution that uses them here answers a question that was not asked.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving that an equation has a solution with the IVT

When to use it: The question says show that the equation has a solution, has a root in an interval, or that $f$ takes the value $N$

1 Move everything to one side and name the function: $h(x) = e^x + x - 3$ .
2 Name the closed interval and justify continuity on ALL of it, with the reason: a sum of an exponential and a polynomial.
3 Compute the two end values EXACTLY and state their signs: $h(0) = -2 < 0$ , $h(1) = e - 2 > 0$ because $e > 2$ .
4 Cite the theorem by name, with $N = 0$ strictly between the two values, and conclude with $c$ in the OPEN interval.
5 Translate back to the original equation, and write at least one, never exactly one.

Concluding sentence

“The function $h(x) = e^x + x - 3$ is continuous on $[0, 1]$ , as the sum of continuous functions. Since $h(0) = -2 < 0$ and $h(1) = e - 2 > 0$ , the Intermediate Value Theorem gives a number $c$ in $(0, 1)$ such that $h(c) = 0$ , that is, $e^c = 3 - c$ .”

The trap: Working with the equation $e^x = 3 - x$ directly, comparing the two sides at the ends without ever defining one continuous function whose sign changes.

Marking: Typically 1 mark for the function, 1 for the continuity on the closed interval, 1 for the two signed values, 1 for the conclusion naming the theorem.

Making a piecewise function continuous

When to use it: The question gives a function by pieces with unknown constants and asks for them

1 Say why each piece is continuous on its OPEN interval, including that a denominator vanishes only at an excluded point.
2 At each seam, write $f(a)$ with the piece that contains the equality sign.
3 Compute each one-sided limit; a piece in the form $\frac{0}{0}$ is simplified before letting $x \to a$ .
4 Write one equation per seam, solve the system, and keep EVERY solution.
5 Check each seam with the values found, and conclude continuous on $\mathbb{R}$ .

Concluding sentence

“Each piece is continuous on its open interval. At $x = 1$ : $\lim_{x\to 1^-} g(x) = 2$ and $g(1) = \lim_{x\to 1^+} g(x) = a + b$ ; at $x = 3$ : $g(3) = 3a + b$ and $\lim_{x\to 3^+} g(x) = 5$ . Hence $a + b = 2$ and $3a + b = 5$ , so $a = \frac{3}{2}$ , $b = \frac{1}{2}$ , and $g$ is continuous on $\mathbb{R}$ .”

The trap: Checking one seam out of two, or writing $f(a)$ with the wrong piece.

Marking: Typically 1 mark per one-sided limit, 1 per equation, 2 for the solution and the check.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Plug the parameters back into both pieces

With the values found, evaluate the two formulas at each seam, after simplifying the zero over zero piece. The two numbers must agree.

With $k = -\frac{1}{2}$ : $k + 3 = \frac{5}{2}$ and $2 - k = \frac{5}{2}$ . If you get $\frac{5}{2}$ and $\frac{3}{2}$ , the equation was set up with the wrong sign.
The interval of the IVT contains no forbidden point

Before citing the theorem, scan the closed interval for zeros of a denominator, negative numbers under a root, or non-positive numbers in a logarithm.

$\tan x$ on $\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$ contains $\frac{\pi}{2}$ : the sign change from $1$ to $-1$ proves nothing.
The extended value equals the simplified formula at the point

After cancelling the common factor, evaluate the simplified formula at a. It must give the value you announced for the extension.

$\frac{x^3 - 8}{x - 2} = x^2 + 2x + 4$ , which gives $12$ at $2$ ; announcing $4$ means the factorization of $x^3 - 8$ was wrong.
Count the pieces of the sign table

Zeros plus points of discontinuity, plus one, is the number of intervals, and each interval needs its own test value.

$\frac{(x + 1)(x - 2)}{x - 3}$ : two zeros and one excluded point, so four intervals and four test values.

The typical problem, taken apart

A function glued at two seams, with two parameters

Find $a$ and $b$ such that $f$ is continuous on $\mathbb{R}$ , where $f(x) = \frac{x^2 + x - 2}{x - 1}$ if $x < 1$ , $f(x) = ax^2 + bx$ if $1 \le x \le 2$ , and $f(x) = \frac{x^2 - 4}{x - 2}$ if $x > 2$ .

No calculator. Every step must be justified as on a MATH 140 midterm.

With

a = -1

and

b = 4

, the arc of parabola on

[1, 2]

picks up the line

y = x + 2

(1, 3)

and hands it back at

(2, 4)

: one unbroken graph.

Step 1

On $(-\infty, 1)$ the first piece is a rational function whose denominator vanishes only at $1$ , excluded; on $(1, 2)$ the second is a polynomial; on $(2, \infty)$ the third is rational with its only zero of the denominator at $2$ , excluded. So $f$ is continuous everywhere except possibly at $1$ and $2$ .

Why

This sentence earns the right to look only at the seams. Without it, the answer proves continuity at two points, not on $\mathbb{R}$ .

Step 2

Seam at $1$ : $f(1) = a + b = \lim_{x\to 1^+} f(x)$ . On the left, the form is $\frac{0}{0}$ : $\frac{x^2 + x - 2}{x - 1} = \frac{(x - 1)(x + 2)}{x - 1} = x + 2 \to 3$ . Condition: $a + b = 3$ .

Why

The value comes from the piece with the equality sign, $1 \le x$ . The left piece is simplified, never evaluated at $1$ , where it does not even apply.

Step 3

Seam at $2$ : $f(2) = 4a + 2b = \lim_{x\to 2^-} f(x)$ . On the right, $\frac{x^2 - 4}{x - 2} = x + 2 \to 4$ . Condition: $4a + 2b = 4$ , that is $2a + b = 2$ .

Why

Every seam is checked, even when the outer pieces look alike: here both simplify to $x + 2$ , which is a coincidence the middle piece has to match twice.

Step 4

Subtract: $(2a + b) - (a + b) = 2 - 3$ , so $a = -1$ , and then $b = 4$ . The middle piece is $-x^2 + 4x$ .

Why

Two unknowns, two seams, one equation each: the count is the first check that the method is complete.

Step 5

Check: $-1 + 4 = 3$ at $1$ , equal to the left limit $3$ ; $-4 + 8 = 4$ at $2$ , equal to the right limit $4$ . So $f$ is continuous at $1$ and at $2$ , hence on $\mathbb{R}$ .

Why

The check is short and catches the usual slip, a sign or a factor $2$ lost in $4a + 2b$ . The figure is the same check drawn.

The conclusion, written out

“Each piece is continuous on its open interval. Continuity at $1$ and at $2$ requires $a + b = 3$ and $4a + 2b = 4$ , so $a = -1$ and $b = 4$ , and with these values $f$ is continuous on $\mathbb{R}$ .”

The classic mistake on this problem: Plugging $x = 1$ into $\frac{x^2 + x - 2}{x - 1}$ , getting $\frac{0}{0}$ and declaring the problem impossible; or writing $f(2)$ with the third piece, which does not contain $x = 2$ .

Learn by heart

• Continuous at $a$ : $f(a)$ defined, $\lim_{x\to a} f(x)$ exists, and they are EQUAL. Three checks, in that order.
• The type of a discontinuity is read from the one-sided LIMITS: equal and finite (removable), finite and different (jump), infinite (infinite).
• Only a removable discontinuity can be repaired, and only by the value of the limit.
• At a seam: left limit $=$ right limit $= f(a)$ , one equation per seam; a $\frac{0}{0}$ piece is simplified, never plugged into.
• Composition: inner limit $b$ , then the OUTER function continuous at $b$ .
• IVT: continuous on the CLOSED $[a, b]$ , $N$ strictly between $f(a)$ and $f(b)$ , then at least one $c$ in $(a, b)$ with $f(c) = N$ .
• Continuous and never zero on an interval: one sign on the whole interval.

Frequently asked questions

How do I show that a function is continuous at a point?

Check three things in order and write each one. First, the function must have a value at the point. Second, the limit at that point must exist, which means the left-hand and right-hand limits are finite and equal. Third, that limit must equal the value. If any of the three fails, the function is discontinuous there, and the one-sided limits tell you which type of discontinuity it is.

What is the difference between a removable and a jump discontinuity?

At a removable discontinuity the limit exists and is finite, but the function is undefined there or has a different value, so redefining one value, the limit, fixes it. At a jump discontinuity the left-hand and right-hand limits are both finite but different, so no single value can make the function continuous.

How do I find the constants that make a piecewise function continuous?

Each piece is continuous on its own open interval, so only the points where the formula changes matter. At each of those points, compute the limit from the left, the limit from the right and the value of the function, and set them equal. That gives one equation per seam. Simplify any piece that gives zero over zero before taking its limit, solve the system, and keep every solution.

When can I use the Intermediate Value Theorem?

Only when the function is continuous on the whole closed interval from a to b, endpoints included. Then every number strictly between f of a and f of b is reached at least once inside the interval. Before citing the theorem, check that no point of the interval makes a denominator zero, a square root negative or a logarithm undefined.

Does the Intermediate Value Theorem tell me how many roots there are?

No. It only guarantees at least one root when a continuous function changes sign on a closed interval. There may be several, and a function that does not change sign may still have roots, like x squared minus one on the interval from minus two to two. To count roots exactly you need another argument, which comes later in the course.

Practise it

Corrected exercises: Continuity and the Intermediate Value Theorem, MATH 140 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Limits and the limit laws Next sheet Infinite limits and asymptotes

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-continuity. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: continuity and the Intermediate Value Theorem (MATH 140)

The essentials

One equation, three checks, and the limits decide the type

Where to look, and what the IVT really says

The rules in table form

Reading a point: the one-sided limits against the value

The mistakes that cost marks

1. Plugging the seam into a piece that gives zero over zero

2. Calling a function continuous because its limit exists

3. Classifying by the first condition that fails

4. Removing an absolute value without cases

5. Keeping one root of a quadratic condition on a parameter

6. Applying the IVT across a point where the function is undefined

7. Invoking the IVT without writing its hypotheses

8. Reading uniqueness, or a location, into the IVT

Which method to choose

Which tool, by the VERB of the question

How the answer is expected to be written

Proving that an equation has a solution with the IVT

Making a piecewise function continuous

Check before you hand in

The typical problem, taken apart

A function glued at two seams, with two parameters

Learn by heart

Frequently asked questions

How do I show that a function is continuous at a point?

What is the difference between a removable and a jump discontinuity?

How do I find the constants that make a piecewise function continuous?

When can I use the Intermediate Value Theorem?

Does the Intermediate Value Theorem tell me how many roots there are?

Corrected exercises: Continuity and the Intermediate Value Theorem, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?