MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: continuity (MATH 203)

Q: How do I find where a function is discontinuous in MATH 203?

List the suspects first: the points excluded from the domain, found on the original formula, and the points where a piecewise formula changes. Everywhere else, sums, products, quotients and compositions of the usual functions are continuous. At each suspect, factor completely, simplify, and compute the one-sided limits: they decide the type.

Q: Why is my discontinuity not removable when a factor cancels?

Because a factor can appear more than once. If the denominator contains x minus c squared and the numerator only once, one copy survives the cancellation, the limit is infinite and no value repairs the function. Count the multiplicity of the factor in the numerator and in the denominator before deciding.

Q: How do I use the Intermediate Value Theorem to show an equation has a solution?

Move everything to one side and name one function. Say why it is continuous on a closed interval, compute its value at both ends and state their signs. If the signs are opposite, the theorem gives at least one solution strictly inside. Then, if asked, locate it with a bisection table on the calculator, in radian mode.

Q: Does a sign change always mean there is a root?

Only for a function that is continuous on the whole closed interval. A function like x plus one over x changes sign between minus one and two without ever being zero, because it is undefined at zero and jumps from minus infinity to plus infinity. Always check the interval for zeros of denominators first.

This sheet is not a summary of section 2.5 of Thomas' Calculus: you already have the course notes. It answers one question only, what makes students lose marks on continuity in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The definition is short, and the marks are lost in the algebra it calls for. At every suspect point the verdict is a limit, and the limit is only as good as the factorization, the complex fraction or the conjugate behind it. Then the Intermediate Value Theorem uses that continuity on a closed interval, and the calculator, in radian mode, locates the root the theorem promised.

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The thread of the chapter

Continuity is decided at the suspect points only, the excluded points and the seams, and there the verdict is a LIMIT that the algebra must reach correctly: factor completely and count what survives the cancellation, clear a complex fraction with its signs, split an absolute value by cases; the IVT then needs that continuity on a CLOSED interval, and the calculator, in radians, only locates what the theorem has guaranteed.

This chapter is part of MATH 203, Calculus I (Concordia)

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
Continuous extension
Continuity of a composite function
Piecewise-defined function
Intermediate Value Theorem
Bisection method

The essentials

Where to look, and the algebra that decides

• $f$ is continuous at $c$ when (1) $f(c)$ exists, (2) $\lim_{x\to c} f(x)$ exists, (3) the two are EQUAL. Check them in this order and write each one.
• Polynomials, rational functions, roots, $|x|$ , $e^x$ , $\ln x$ , $\sin$ , $\cos$ are continuous on their domains. The only suspects are the EXCLUDED points and the SEAMS of a piecewise formula.
• At a suspect point, factor numerator and denominator COMPLETELY (grouping, $a^3 - b^3$ , leading coefficient), cancel, then look at what is left.
• If a factor $x - c$ is left in the denominator, the discontinuity is infinite, even after a cancellation: $(x - c)^2$ loses only one copy.
• A complex fraction is cleared by multiplying top and bottom by the common denominator, and $\frac{c - x}{x - c} = -1$ , never $1$ .
• The type is read from the one-sided limits: equal and finite (removable), finite and different (jump), infinite (infinite).

Same numerator

x^2 - x - 2

: over

x - 2

the factor cancels and leaves a hole at

(2, 3)

; over

(x - 2)^2

one copy survives and the graph escapes to infinity.

The domain is decided on the ORIGINAL formula, before any cancellation: $\frac{(x + 3)(x - 2)}{(x + 3)(x - 2)(x + 2)}$ is undefined at $-3$ and $2$ even though it simplifies to $\frac{1}{x + 2}$ .

The IVT, then the calculator

• IVT: $f$ continuous on the CLOSED interval $[a, b]$ and $y_0$ between $f(a)$ and $f(b)$ give some $c$ in $[a, b]$ with $f(c) = y_0$ .
• For an equation, move everything to ONE side and name one function: $\cos x = x^2$ becomes $g(x) = \cos x - x^2 = 0$ .
• The IVT gives AT LEAST one solution: never the number of solutions, never where exactly.
• Bisection: evaluate at the midpoint, keep the half where the sign changes; after $n$ steps the interval has length $\frac{b - a}{2^n}$ .
• Every trigonometric value is computed in RADIAN mode. A cosine that stays near $1$ for every input is the signature of degree mode.

A marker gives the IVT marks for three written sentences: the function and the closed interval, WHY it is continuous there, the two end values with their signs. The table of bisection comes after, never instead.

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.

What survives the cancellation decides the type

Factor completely near the suspect point $c$ , cancel what can be cancelled, and read the verdict from what is left. The verdict never comes from the fact that the denominator vanishes.

Near c, after factoring	Limit at c	Verdict
$\frac{(x - c)A}{(x - c)B}$ , $B(c) \ne 0$	$\frac{A(c)}{B(c)}$	removable
Example: $\frac{2x^2 - 5x + 3}{x^3 - 1} = \frac{2x - 3}{x^2 + x + 1} \to -\frac{1}{3}$ at $1$ .
$\frac{(x - c)A}{(x - c)^2 B}$	$\pm\infty$	infinite
Example: $\frac{x^2 - 1}{x^2 - 2x + 1} = \frac{x + 1}{x - 1}$ : numerator near $2$ , denominator near $0$ at $1$ .
$\frac{A}{(x - c)B}$ , $A(c) \ne 0$	$\pm\infty$	infinite
Example: $\frac{x^2 + 2x - 15}{x^2 - 9}$ at $-3$ : the numerator is $9 - 6 - 15 = -12$ .
$\frac{\|x - c\|}{x - c} A$	$-A(c)$ and $A(c)$	jump
Example: $\frac{\|x - 3\|(x + 1)}{x - 3}$ at $3$ : $-4$ from the left, $4$ from the right.

Only the first line has a continuous extension, and its value is the limit $\frac{A(c)}{B(c)}$ .

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. Cancelling one copy of a repeated factor

the whole question

What not to write

“ $\frac{x^2 - 1}{x^2 - 2x + 1}$ : the factor $x - 1$ cancels, so the discontinuity at $1$ is removable.”

What to write

“ $\frac{(x - 1)(x + 1)}{(x - 1)^2} = \frac{x + 1}{x - 1}$ for $x \ne 1$ : a factor $x - 1$ remains below, the limit is infinite, and no value $q(1)$ makes $q$ continuous.”

Why: Removable means that the limit EXISTS. Count the multiplicity of the factor before cancelling: what is left decides.

2. Losing the sign when clearing a complex fraction

2 marks

What not to write

“ $\frac{\frac{1}{x} - \frac{1}{3}}{x - 3} = \frac{3 - x}{3x(x - 3)} = \frac{1}{3x}$ , so the limit at $3$ is $\frac{1}{9}$ .”

What to write

“ $\frac{3 - x}{3x(x - 3)} = \frac{-(x - 3)}{3x(x - 3)} = -\frac{1}{3x}$ , so the limit at $3$ is $-\frac{1}{9}$ .”

Why: $3 - x$ and $x - 3$ are opposites. The wrong answer looks perfectly reasonable, which is why a calculator value on each side, $h(3.001) \approx -0.1111$ , is worth the ten seconds.

3. Stopping a factorization by grouping halfway

2 to 3 marks

What not to write

“ $x^3 + 3x^2 - 4x - 12 = x^2(x + 3) - 4(x + 3)$ has no factor in common with $(x + 3)(x - 2)$ , so each zero of the denominator is an asymptote.”

What to write

“ $x^2(x + 3) - 4(x + 3) = (x + 3)(x^2 - 4) = (x + 3)(x - 2)(x + 2)$ . The discontinuities at $-3$ and $2$ are removable; only $-2$ is infinite.”

Why: $x^2(x + 3) - 4(x + 3)$ is still a SUM: nothing can be cancelled until the common factor $x + 3$ has been pulled out, and $x^2 - 4$ still factors.

4. Removing an absolute value the same way on both sides

2 marks

What not to write

“ $\frac{|x - 1| - 2}{x^2 - 2x - 3} = \frac{x - 3}{(x - 3)(x + 1)} = \frac{1}{x + 1}$ , so there is an asymptote at $x = -1$ .”

What to write

“Near $-1$ , $x - 1 < 0$ and $|x - 1| = 1 - x$ : $\frac{-(x + 1)}{(x - 3)(x + 1)} = -\frac{1}{x - 3} \to \frac{1}{4}$ . The discontinuity at $-1$ is removable, like the one at $3$ .”

Why: $|x - 1| = x - 1$ only for $x \ge 1$ . The sign of the quantity inside the bars is read AT the point studied, and it decides which formula applies there.

5. Copying the bad point of the outer function in a composition

1 mark

What not to write

“ $f(u) = \frac{1}{u - 1}$ fails at $1$ , so $f(x^2) = \frac{1}{x^2 - 1}$ is discontinuous at $x = 1$ .”

What to write

“ $f \circ g$ fails where $g(x) = 1$ , that is $x^2 = 1$ : at $x = -1$ and $x = 1$ .”

Why: The question is which $x$ are SENT to the bad point, and that is an equation to solve, $g(x) = 1$ , with all its solutions.

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

the whole question

What not to write

“ $w(x) = x + \frac{1}{x}$ has $w(-1) = -2 < 0$ and $w(2) = \frac{5}{2} > 0$ , so $w$ has a zero in $(-1, 2)$ .”

What to write

“ $w$ is undefined at $0$ , inside $[-1, 2]$ , so the IVT does not apply. In fact $x + \frac{1}{x} = \frac{x^2 + 1}{x}$ is never $0$ .”

The curve starts below the axis at

x = -1

and ends above it at

x = 2

, but it jumps at

x = 0

instead of crossing: no zero, and no contradiction with the IVT.

Why: The sign changes by jumping through infinity, not by crossing the axis. Scan the closed interval for zeros of denominators BEFORE quoting the theorem.

7. Running a bisection in degree mode

the whole question

What not to write

“ $g(x) = \sin x - \frac{x}{2}$ : $g(1.5) \approx -0.7238$ and $g(2) \approx -0.9651$ , no sign change, so $\sin x = \frac{x}{2}$ has no solution in $(1.5, 2)$ .”

What to write

“In radians, $g(1.5) \approx 0.2475 > 0$ and $g(2) \approx -0.0907 < 0$ : by the IVT there is a solution in $(1.5, 2)$ .”

Why: Calculus is done in radians: in degrees, $\sin 1.5$ is the sine of one and a half degrees, $0.0262$ . A sine or cosine that barely moves between two inputs is the alarm.

8. Reading a failed hypothesis as a proof of the opposite

1 to 2 marks

What not to write

“ $f(0) < 0 < f(2)$ , but $f$ is discontinuous at $1$ , so $f$ has no zero in $(0, 2)$ .”

What to write

“ $f$ is not continuous on $[0, 2]$ , so the IVT says nothing: a zero may or may not exist. For instance $x - \frac{1}{2}$ for $x < 1$ and $x$ for $x \ge 1$ vanishes at $\frac{1}{2}$ .”

Why: A theorem whose hypothesis fails gives no conclusion at all, not the opposite one. Only a direct computation or a counterexample settles the question.

Which method to choose

Which gesture at a suspect point, by the FORM of the formula

Plug the suspect point in first: a non-zero number over zero is settled at once, a zero over zero calls for one of the gestures below

If non-zero numerator over a zero denominator → infinite discontinuity, stop: no extension exists

Example: $\frac{x^2 + 2x - 15}{x^2 - 9}$ at $-3$ : $\frac{-12}{0}$
If polynomial over polynomial, form $\frac{0}{0}$ → factor completely (grouping, difference of cubes, leading coefficient), cancel, count what is left

Example: $\frac{2x^2 - 5x + 3}{x^3 - 1} \to -\frac{1}{3}$ at $1$
If a difference of square roots → multiply top and bottom by the conjugate, keep the parentheses

Example: $\frac{\sqrt{3 + x} - \sqrt{3 - x}}{x} \to \frac{1}{\sqrt 3}$ at $0$
If fractions inside a fraction → multiply top and bottom by the common denominator of the small fractions

Example: $\frac{\frac{2}{x} + 2}{x + 1} = \frac{2}{x} \to -2$ at $-1$
If an absolute value that vanishes at the point → two computations, one per side, with the sign of the inside

Example: $\frac{|x - 1| - 2}{x^2 - 2x - 3} \to \frac{1}{4}$ at $-1$ , with $|x - 1| = 1 - x$
If trigonometric $\frac{0}{0}$ → an identity of chapter 1, then cancel

Example: $\frac{\sin^2 x}{1 - \cos x} = 1 + \cos x \to 2$ at $0$
If a negative exponent → rewrite it as a fraction before anything else

Example: $\frac{x^{-1} - x}{x - 1} = -\frac{x + 1}{x} \to -2$ at $1$

No derivative and no L'Hôpital's Rule in this chapter: every limit here is reached by algebra alone, and that is what the marker checks.

Which tool, by the VERB of the question

The verb of the question picks the tool before any computation

If where is f continuous? → domain from the original formula, then the seams, then the endpoints one-sided

Example: $\frac{\sqrt{6 - 2x}}{x^2 - x - 2}$ : $(-\infty, -1)$ , $(-1, 2)$ , $(2, 3]$
If find the constant(s) that make f continuous → one equation per seam, left limit = right limit = value, keep every solution

Example: $9 - c^2 = 3c - 1$ : $c = 2$ or $c = -5$
If show that the equation has a solution → one function, a closed interval, continuity, two signs, the IVT

Example: $\cos x - x^2$ : $1$ at $0$ , $\cos 1 - 1 < 0$ at $1$
If locate the solution to within a given length → bisection table in radians, one line per midpoint

Example: $[0, 1]$ after four steps: $[0.8125, 0.875]$

If none of the verbs matches, the question is a definition question: write the three conditions at the point.

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, then locating it

When to use it: The question says show that the equation has a solution in an interval, then approximate it or find an interval of a given length

1 Move everything to one side and name the function: $h(x) = e^{-x} - x$ .
2 Name the closed interval and justify continuity on ALL of it, with the reason: a difference of an exponential and a polynomial.
3 Compute the two end values and state their signs: $h(0) = 1 > 0$ , $h(1) = e^{-1} - 1 \approx -0.6321 < 0$ .
4 Quote the theorem by name, with $0$ between the two values, and conclude with $c$ in the OPEN interval.
5 Then the table: $h(0.5) \approx 0.1065 > 0$ , keep $[0.5, 1]$ ; $h(0.75) \approx -0.2776 < 0$ , keep $[0.5, 0.75]$ .

Concluding sentence

“The function $h(x) = e^{-x} - x$ is continuous on $[0, 1]$ , as a difference of continuous functions. Since $h(0) = 1 > 0$ and $h(1) = e^{-1} - 1 < 0$ , the Intermediate Value Theorem gives a number $c$ in $(0, 1)$ such that $h(c) = 0$ , that is, $e^{-c} = c$ . By bisection, $c$ lies in $(0.5, 0.75)$ .”

The trap: Comparing $e^{-x}$ and $x$ at the two ends without ever defining ONE continuous function whose sign changes, or giving the bisection table without the theorem that makes it meaningful.

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, then the table.

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 $f(c)$ with the piece that contains the equality sign.
3 Compute each one-sided limit; a piece in the form $\frac{0}{0}$ is simplified by the gesture its form calls for, then $x \to c$ .
4 Write one equation per seam, solve the system, and keep EVERY solution.
5 Plug the values back into both pieces at each seam, then conclude continuous on the whole domain.

Concluding sentence

“Each piece is continuous on its open interval. At $x = 3$ : $\lim_{x\to 3^-} f(x) = \frac{7}{2}$ and $f(3) = \lim_{x\to 3^+} f(x) = 3a + b$ ; at $x = 5$ : $f(5) = 5a + b$ and $\lim_{x\to 5^+} f(x) = \frac{9}{2}$ . Hence $3a + b = \frac{7}{2}$ and $5a + b = \frac{9}{2}$ , so $a = \frac{1}{2}$ , $b = 2$ , and $f$ is continuous on $\mathbb{R}$ .”

The trap: Evaluating a $\frac{0}{0}$ piece at its own seam and declaring the problem impossible, or checking one seam out of two.

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.

A calculator value on each side of the point

Evaluate the ORIGINAL formula at c - 0.001 and c + 0.001, in radian mode. Both values must be close to the limit you announced.

For $\frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ : $h(3.001) \approx -0.1111$ . If you announced $+\frac{1}{9}$ , the sign was lost in the complex fraction.
Plug the parameters back into both pieces

With each value found, evaluate the two formulas at the seam. The two numbers must agree, for EVERY value kept.

With $c = -5$ : $9 - 25 = -16$ and $-15 - 1 = -16$ . If you found $c = 5$ , the check gives $-16$ and $14$ and exposes the sign slip.
The IVT interval contains no forbidden point

Before quoting the theorem, scan the closed interval for zeros of denominators, negative numbers under a root, non-positive numbers in a logarithm, and odd multiples of pi over 2 for a tangent.

$\tan x - x$ on $[1, 2]$ contains $\frac{\pi}{2} \approx 1.5708$ : the sign change from $0.5574$ to $-4.1850$ proves nothing.
The bisection interval has the right length and opposite end signs

After n steps from an interval of length L, the final interval has length L over 2 to the n, and the function has opposite signs at its two ends.

Four steps from $[0, 1]$ : length $\frac{1}{16} = 0.0625$ , and $\cos x - x^2$ is $+0.0275$ at $0.8125$ , $-0.1246$ at $0.875$ .

The typical problem, taken apart

A function glued at two seams, with trinomials to factor

Find $a$ and $b$ such that $f$ is continuous on $\mathbb{R}$ , where $f(x) = \frac{2x^2 - 5x - 3}{2x - 6}$ if $x < 3$ , $f(x) = ax + b$ if $3 \le x \le 5$ , and $f(x) = \frac{2x^2 - 11x + 5}{2x - 10}$ if $x > 5$ .

Every step must be justified as on a MATH 203 midterm.

With

a = \frac{1}{2}

and

b = 2

, the middle segment meets

y = x + \frac{1}{2}

\left(3, \frac{7}{2}\right)

and

y = x - \frac{1}{2}

\left(5, \frac{9}{2}\right)

: one unbroken graph.

Step 1

On $(-\infty, 3)$ the first piece is rational and its denominator $2x - 6$ vanishes only at $3$ , excluded; on $(3, 5)$ the second is a polynomial; on $(5, \infty)$ the third is rational with its only zero of the denominator at $5$ , excluded. So $f$ is continuous everywhere except possibly at $3$ and $5$ .

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 $3$ : $f(3) = 3a + b = \lim_{x\to 3^+} f(x)$ . On the left the form is $\frac{0}{0}$ . Factor with the leading coefficient: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$ , and $2x - 6 = 2(x - 3)$ , so $f(x) = \frac{2x + 1}{2} \to \frac{7}{2}$ . Condition: $3a + b = \frac{7}{2}$ .

Why

The factorization is checked by expanding, $2x^2 - 6x + x - 3$ ; forgetting the $2$ of $2x - 6$ gives $7$ instead of $\frac{7}{2}$ , and the whole system follows it.

Step 3

Seam at $5$ : $f(5) = 5a + b = \lim_{x\to 5^-} f(x)$ . On the right, $2x^2 - 11x + 5 = (2x - 1)(x - 5)$ and $2x - 10 = 2(x - 5)$ , so $f(x) = \frac{2x - 1}{2} \to \frac{9}{2}$ . Condition: $5a + b = \frac{9}{2}$ .

Why

Every seam is checked. The value $f(5)$ comes from the middle piece, which contains $x = 5$ ; the right piece is never evaluated at $5$ .

Step 4

Subtract: $(5a + b) - (3a + b) = \frac{9}{2} - \frac{7}{2}$ , so $2a = 1$ and $a = \frac{1}{2}$ ; then $b = \frac{7}{2} - \frac{3}{2} = 2$ .

Why

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

Step 5

Check: $\frac{3}{2} + 2 = \frac{7}{2}$ at $3$ and $\frac{5}{2} + 2 = \frac{9}{2}$ at $5$ , equal to the limits. So $f$ is continuous at $3$ and at $5$ , hence on $\mathbb{R}$ .

Why

The check is short and catches the usual slips, a factor $2$ or a sign. The figure is the same check drawn.

The conclusion, written out

“Each piece is continuous on its open interval. Continuity at $3$ and at $5$ requires $3a + b = \frac{7}{2}$ and $5a + b = \frac{9}{2}$ , so $a = \frac{1}{2}$ and $b = 2$ , and with these values $f$ is continuous on $\mathbb{R}$ .”

The classic mistake on this problem: Cancelling $x - 3$ in the numerator without first writing $2x - 6 = 2(x - 3)$ , which leaves nothing to cancel it against, or writing the left limit as $7$ after losing the factor $2$ .

Learn by heart

• Continuous at $c$ : $f(c)$ exists, $\lim_{x\to c} f(x)$ exists, and they are EQUAL. Three checks, in that order.
• Suspects: the excluded points and the seams. Everywhere else the continuity theorems decide.
• Factor completely, cancel, and read what is LEFT: a surviving $x - c$ below means infinite.
• $\frac{c - x}{x - c} = -1$ ; $x^{-1} = \frac{1}{x}$ ; $|x| = -x$ for $x < 0$ .
• At a seam: left limit $=$ right limit $= f(c)$ , one equation per seam, every solution kept and checked.
• IVT: continuous on the CLOSED $[a, b]$ , $y_0$ between $f(a)$ and $f(b)$ , then at least one $c$ with $f(c) = y_0$ .
• Bisection in RADIANS: after $n$ steps the interval has length $\frac{b - a}{2^n}$ .

Frequently asked questions

How do I find where a function is discontinuous in MATH 203?

List the suspects first: the points excluded from the domain, found on the original formula, and the points where a piecewise formula changes. Everywhere else, sums, products, quotients and compositions of the usual functions are continuous. At each suspect, factor completely, simplify, and compute the one-sided limits: they decide the type.

Why is my discontinuity not removable when a factor cancels?

Because a factor can appear more than once. If the denominator contains x minus c squared and the numerator only once, one copy survives the cancellation, the limit is infinite and no value repairs the function. Count the multiplicity of the factor in the numerator and in the denominator before deciding.

How do I use the Intermediate Value Theorem to show an equation has a solution?

Move everything to one side and name one function. Say why it is continuous on a closed interval, compute its value at both ends and state their signs. If the signs are opposite, the theorem gives at least one solution strictly inside. Then, if asked, locate it with a bisection table on the calculator, in radian mode.

Does a sign change always mean there is a root?

Only for a function that is continuous on the whole closed interval. A function like x plus one over x changes sign between minus one and two without ever being zero, because it is undefined at zero and jumps from minus infinity to plus infinity. Always check the interval for zeros of denominators first.

Practise it

Corrected exercises: Continuity, MATH 203 at Concordia

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 involving infinity and asymptotes Next sheet The derivative at a point and as a function

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-continuity. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: continuity (MATH 203)

The essentials

Where to look, and the algebra that decides

The IVT, then the calculator

The rules in table form

What survives the cancellation decides the type

The mistakes that cost marks

1. Cancelling one copy of a repeated factor

2. Losing the sign when clearing a complex fraction

3. Stopping a factorization by grouping halfway

4. Removing an absolute value the same way on both sides

5. Copying the bad point of the outer function in a composition

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

7. Running a bisection in degree mode

8. Reading a failed hypothesis as a proof of the opposite

Which method to choose

Which gesture at a suspect point, by the FORM of the formula

Which tool, by the VERB of the question

How the answer is expected to be written

Proving that an equation has a solution, then locating it

Making a piecewise function continuous

Check before you hand in

The typical problem, taken apart

A function glued at two seams, with trinomials to factor

Learn by heart

Frequently asked questions

How do I find where a function is discontinuous in MATH 203?

Why is my discontinuity not removable when a factor cancels?

How do I use the Intermediate Value Theorem to show an equation has a solution?

Does a sign change always mean there is a root?

Corrected exercises: Continuity, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?