Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: limits and continuity (201-103)

The chapter looks easy because there is little computation in it. It is in fact the one where papers lose the most reasoning marks, because the question asked is almost never the one that gets answered: you are asked where the function is heading, and you answer what it equals.

This sheet takes the eight mistakes that come back every term in 201-103, puts a price on each, and gives for each the exact sentence to copy onto your paper. It also gives the two write-ups the course expects, the one for the Intermediate Value Theorem and the one for making a benefit schedule continuous.

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The thread of the chapter

A limit describes what the function does AROUND the point, never AT the point: almost every mistake in this chapter comes from answering the question about the value when the question asked was about the trend.

This chapter is part of 201-103, Differential Calculus for Social Science

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

One-sided limits
Continuity at a point
Hole (removable discontinuity)
Jump discontinuity
Indeterminate forms
Conjugate expression
Vertical asymptote
Horizontal asymptote
Intermediate Value Theorem
Piecewise-defined function
Absolute and relative gap

The essentials

Three different questions at a single point

• $\lim_{x\to a^{-}}f(x)$ : where the function is heading as you arrive FROM THE LEFT.
• $\lim_{x\to a^{+}}f(x)$ : where it is heading as you arrive FROM THE RIGHT.
• $f(a)$ : what it EQUALS exactly at the point. This value may not exist, or may exist and differ from the other two.
• The limit $\lim_{x\to a}f(x)$ exists if and only if both one-sided limits exist AND are equal.
• Three discontinuities, and you must NAME it: a hole if the limit exists but differs from $f(a)$ , and it is removable; a jump if the one-sided limits differ, and it is not removable; infinite if there is a vertical asymptote.

On the left the curve aims at

2.5

from both sides while the filled dot is at

4.6

; in the middle the two sides do not aim at the same number; on the right there is nothing left to aim at.

Naming the condition that fails is almost always worth a mark of its own: “the limit exists, but $f(4)\neq 3$ ” earns more than “ $f$ is not continuous”.

The three conditions for continuity at $a$

• $f(a)$ exists, that is, $a$ belongs to the domain.
• $\lim_{x\to a}f(x)$ exists, that is, the two one-sided limits are equal.
• The two coincide: $\lim_{x\to a}f(x)=f(a)$ .
• The three are checked IN THIS ORDER, and you stop at the first one that fails, naming it.
• A polynomial is continuous everywhere. A rational function is continuous wherever its denominator does not vanish. These two sentences justify continuity in nine questions out of ten.

“ $f$ is a polynomial, so it is continuous on $[a, b]$ ” is a complete and sufficient justification. “ $f$ is clearly continuous” is not one.

Asymptotes, and what they really forbid

• VERTICAL asymptote at $a$ : the denominator tends to $0$ and the numerator does NOT tend to $0$ . The sign of the denominator on each side decides between $+\infty$ and $-\infty$ .
• HORIZONTAL asymptote: a behaviour at infinity, so a long-term trend. The curve is perfectly free to cross it along the way.
• The average cost $\dfrac{F}{n}+v$ has a vertical asymptote at $0$ and a horizontal asymptote $y=v$ : economies of scale are real, and bounded by the variable cost.
• An ABSOLUTE gap that tends to infinity is compatible with a RELATIVE gap that tends to zero. The two do not measure the same thing, and the statement always says which one it wants.

The average cost

\frac{2{,}000}{n}+14

shoots off to infinity as

n\to 0^{+}

(the vertical axis is an asymptote) and falls toward

14

dollars without reaching it: real economies of scale, bounded.

A vertical asymptote is a boundary of the domain; a horizontal asymptote is not. Confusing the two is the most profitable trap on the exam.

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.

Which limit, according to the form obtained after substitution

Substitute first. The form obtained decides everything, and the last column gives the answer. The two red cells are not answers: they are orders to rewrite the expression.

Form obtained	What to look at	Limit
$\dfrac{P}{Q}$ , $\deg P<\deg Q$	the degrees	$0$
Example: $\dfrac{3x+5}{x^{2}+1}$ is about $0.0305$ at $x=100$ , and tends to $0$ .
$\dfrac{P}{Q}$ , $\deg P=\deg Q$	the leading coefficients	their ratio
Example: $\dfrac{6x^{3}+x}{2x^{3}-5}$ is about $3.0001$ at $x=100$ , and tends to $3$ .
$\dfrac{P}{Q}$ , $\deg P>\deg Q$	the sign of the leading term	$+\infty$ or $-\infty$
Example: $\dfrac{x^{3}}{x^{2}+1}$ is about $99.99$ at $x=100$ , and tends to $+\infty$ .
$\dfrac{0}{0}$	a common factor	to cancel settles nothing
Example: $\lim_{x\to 3}\dfrac{x^{2}-9}{x-3}=6$ , after factoring out $(x-3)$ . Same form, other result: $\lim_{x\to 3}\dfrac{(x-3)^{2}}{x-3}=0$ : same form, different answer. What to do: Factor out $(x-a)$ on top AND bottom, then cancel before substituting again.
$\infty-\infty$	the conjugate expression	to transform settles nothing
Example: $\lim_{x\to+\infty}\left(\sqrt{x^{2}+6x}-x\right)=3$ . Same form, other result: $\lim_{x\to+\infty}\left(\sqrt{x^{2}+2x}-x\right)=1$ : same form, different answer. What to do: Multiply and divide by $\sqrt{x^{2}+bx}+x$ , then cancel by $x$ .

The first three rows are read at a glance from the degrees: this is the most profitable calculation of the chapter, and it requires no transformation.

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. Confusing the limit and the value of the function

2 marks, and the error comes back in the derivative chapter

What not to write

“ $f(4)$ does not exist, so $\lim_{x\to 4}f(x)$ does not exist.”

What to write

“ $f$ is not defined at $4$ , but $\lim_{x\to 4}f(x)=4$ : the limit never looks at the point itself.”

f(x)=\frac{x^{2}-4x}{x-4}

coincides with

y=x

except at

4

(open dot). From both sides, the curve heads toward height

4

: the limit is

4

, even though

f(4)

does not exist.

Why: The derivative is precisely the limit of a quotient that does NOT exist at the point. A student who confuses the two objects here will not be able to understand the definition of the derivative three weeks from now.

2. Declaring a vertical asymptote as soon as the denominator vanishes

3 marks, and a curve study that is entirely wrong

What not to write

“The denominator of $\dfrac{x^{2}-9}{x-3}$ vanishes at $3$ , so the line $x=3$ is a vertical asymptote.”

What to write

“At $3$ , the numerator ALSO vanishes: the factor $(x-3)$ cancels and the limit is $6$ . There is a hole, not an asymptote.”

Both denominators vanish, and something completely different happens: on the left the line continues on both sides of the open circle, on the right the curve heads off to infinity.

Why: An asymptote requires the denominator to tend to zero and the numerator NOT to. Testing both takes five seconds and avoids drawing an infinite branch where there is nothing.

3. Treating $\dfrac{0}{0}$ as a result

the whole question, since the announced answer is a wrong number

What not to write

“ $\lim_{x\to 3}\dfrac{x^{2}-9}{x-3}=\dfrac{0}{0}=0$ .”

What to write

“Substitution gives the indeterminate form $\dfrac{0}{0}$ : I factor, $\dfrac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq 3$ , so the limit is $6$ .”

Why: $\dfrac{0}{0}$ is not a number, it is a report that substitution failed. It signals that a common factor remains to be cancelled, nothing more.

4. Taking the horizontal asymptote for a barrier

1 interpretation mark, and a wrong shape of the curve when sketching

What not to write

“The limit at infinity is $3$ , so the curve always stays below $y=3$ .”

What to write

“The curve tends to $3$ at infinity, but nothing prevents it from crossing $y=3$ several times before settling there.”

The curve crosses the dashed line four times before sticking to it: tending to

3

forbids neither equalling

3

nor passing above it.

Why: A horizontal asymptote describes a LONG-TERM behaviour, not a constraint. An economic series that converges to its trend has very often overshot it along the way.

5. Concluding uniqueness with the Intermediate Value Theorem

1 mark, deducted every time

What not to write

“By the IVT, there is a unique $c$ such that $f(c)=0$ .”

What to write

“By the IVT, there is AT LEAST one $c\in(0, 10)$ such that $f(c)=0$ . Since $f$ is also strictly increasing on this interval, this $c$ is unique.”

Continuous, negative at the start, positive at the end: the IVT guarantees a zero for each. Only strict increase (left) guarantees ONLY ONE; on the right, three.

Why: The theorem gives an existence, never a uniqueness. Uniqueness always comes from an ADDITIONAL argument, strict monotonicity, and this argument must be written.

6. Applying the IVT without checking its hypotheses

2 marks out of 4: the marking scheme pays for the hypotheses as much as for the conclusion

What not to write

“ $f(0)=-3$ and $f(10)=77$ , so there is a $c$ such that $f(c)=0$ .”

What to write

“ $f$ is a polynomial, so it is continuous on the CLOSED interval $[0, 10]$ . Now $f(0)=-3<0<77=f(10)$ . By the IVT, there is a $c\in(0, 10)$ such that $f(c)=0$ .”

Negative at

0

, positive at

10

, and yet never zero: the function JUMPS at

x=5

. Without continuity on

[0, 10]

, the IVT says nothing.

Why: Without continuity, the theorem is false: a function that jumps over $0$ goes from negative to positive without ever vanishing. That is exactly what a discontinuous benefit schedule does.

7. Making a schedule continuous with a single equality

1 rigour mark, sometimes 2 when the statement asks for a justification

What not to write

“At the junction $x=3000$ , the second formula must equal $450$ , so $k=900{,}000$ .”

What to write

“The three conditions are checked: $A(3000)=450$ by the first formula, $\lim_{x\to 3000^{+}}A(x)=\dfrac{k}{2000}$ , and equating the two forces $k=900{,}000$ .”

Why: The result is the same, the write-up is not. The marking scheme pays for writing out the three conditions, because that is what proves you know what continuity means.

8. Confusing the absolute gap and the relative gap

2 interpretation marks, and this is the reasoning question of the chapter

What not to write

“The gap between the two budgets tends to infinity, so they diverge more and more.”

What to write

“The ABSOLUTE gap $50t$ tends to infinity, but the ratio $\dfrac{t^{2}+50t}{t^{2}}$ tends to $1$ : in proportion, the two budgets converge.”

On the left,

t^{2}

and

t^{2}+50t

drift apart by

50t

, without bound. On the right, their ratio tends to

1

: in proportion, the two budgets converge.

Why: This is the distinction specific to the social sciences, and it turns up everywhere: an income gap that grows in dollars can shrink in percentage. The statement always says which one it wants.

9. Forgetting that the SIGN of the denominator decides between the two infinities

1.5 marks, and a curve drawn on the wrong side

What not to write

“ $\lim_{x\to 2}\dfrac{1}{x-2}=+\infty$ .”

What to write

“ $\lim_{x\to 2^{-}}\dfrac{1}{x-2}=-\infty$ and $\lim_{x\to 2^{+}}\dfrac{1}{x-2}=+\infty$ : the limit at $2$ does not exist, and the line $x=2$ is a vertical asymptote.”

To the left of

2

, the denominator is negative and small: the curve plunges toward

-\infty

. To the right, it is positive: the curve climbs toward

+\infty

. A single two-sided limit cannot sum this up.

Why: Near an asymptote, the two sides are almost always opposite. Writing a single two-sided limit amounts to asserting that they are equal, which is false one time in two.

Which method to choose

Which technique, according to the form obtained after substitution

ALWAYS substitute first. Nine times out of ten substitution gives the answer directly; the rest of the time, the form obtained names the technique.

If substitution gives a number → that is the limit, stop there

Example: $\lim_{x\to 1}\dfrac{x+3}{x^{2}+1}=\dfrac{4}{2}=2$
If form $\dfrac{0}{0}$ with polynomials → factor out $(x-a)$ on top and bottom, then cancel

Example: $\dfrac{x^{2}-9}{x-3}\to 6$

the factor $(x-a)$ is ALWAYS present on both sides, otherwise the form would not be $\dfrac{0}{0}$
If form $\dfrac{0}{0}$ with a square root → multiply top and bottom by the conjugate expression

Example: $\dfrac{\sqrt{x+4}-2}{x}\to\dfrac{1}{4}$
If form $\infty-\infty$ with a root → multiply and divide by the conjugate, then compare degrees

Example: $\sqrt{x^{2}+6x}-x\to 3$
If form $\dfrac{\infty}{\infty}$ → compare the DEGREES, without expanding anything

Example: $\dfrac{6x^{3}+x}{2x^{3}-5}\to 3$
If a nonzero number over $0$ → vertical asymptote: study the SIGN of the denominator on each side

Example: $\dfrac{1}{x-2}$ : $-\infty$ on the left, $+\infty$ on the right
If a piecewise-defined function, and $a$ is a junction → compute the two one-sided limits SEPARATELY, each with its own formula

Example: $A(3000^{-})=450$ and $A(3000^{+})=\dfrac{k}{2000}$

If no branch applies, it is almost always because substitution was not done. L'Hospital's rule is NOT in the program of this course: a paper that uses it receives no marks for the question.

Which theorem, according to the verb in the statement

Read the verb before computing. It tells you which theorem is expected and which concluding sentence will be paid.

If “show that there is a value such that” → IVT: continuity on a CLOSED interval, then $k$ between the two images

Example: show that a threshold of $0$ is crossed between $0$ and $10$
If “show that there is a UNIQUE value” → IVT for existence, THEN strict monotonicity for uniqueness

Example: two arguments, so two sentences

monotonicity is justified by the sign of the derivative, or by the shape of the function
If “is the function continuous at” → the three conditions, in order, and name the one that fails

Example: does $f(4)$ exist, does the limit exist, are they equal
If “determine the parameter so that” → write the equality of the two one-sided limits and solve

Example: $k$ such that the schedule does not jump at $3000$
If “what becomes of the model in the long run” → limit at infinity, then interpretation in words and units

Example: the average cost tends to the variable cost

A statement that asks for neither existence, nor uniqueness, nor continuity does not ask for a theorem: it asks for a calculation, and a theorem sentence earns nothing there.

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.

Writing up an application of the Intermediate Value Theorem

When to use it: As soon as the statement asks you to show that a threshold is reached, that an equation has a solution, or that a quantity passes through a given value.

1 Name the function and the CLOSED interval you are working on.
2 Justify continuity with a REASON: “ $f$ is a polynomial, so it is continuous on $[0, 10]$ ”.
3 Compute the two images at the endpoints and write them down.
4 Check that the target value lies between the two, with the inequalities written out.
5 State the conclusion citing the theorem, with “AT LEAST one”.
6 If the statement asks for uniqueness, add strict monotonicity as a SEPARATE argument.

Concluding sentence

“The function $f$ is a polynomial, so it is continuous on $[0, 10]$ . Now $f(0)=-3<0$ and $f(10)=77>0$ , so $0$ lies between $f(0)$ and $f(10)$ . By the Intermediate Value Theorem, there is at least one $c\in(0, 10)$ such that $f(c)=0$ .”

The trap: Skipping the justification of continuity because it seems obvious. It is the most expensive hypothesis of the theorem and the one the grader looks for first.

Marking: 1 mark for the justified continuity, 1 mark for the two images, 1 mark for the bracketing, 1 mark for the conclusion citing the theorem.

Making a piecewise schedule continuous

When to use it: As soon as a statement gives a function in two formulas with a parameter and asks that there be no jump at the junction.

1 Locate the junction point and say that it is THERE, and nowhere else, that continuity is in question.
2 Compute $f(a)$ with the formula that includes $a$ , and say so.
3 Compute the left-hand limit with the left formula, then the right-hand limit with the right formula, separately.
4 Set the two one-sided limits equal and solve for the parameter.
5 Check that the value found does give back $f(a)$ , and conclude with a sentence.
6 Interpret: say what a discontinuity would have meant in the context of the statement.

Concluding sentence

“For $x\le 3000$ , $A(3000)=900-0.15\times 3000=450$ . For $x>3000$ , $\lim_{x\to 3000^{+}}\dfrac{k}{x-1000}=\dfrac{k}{2000}$ . Continuity requires $\dfrac{k}{2000}=450$ , so $k=900{,}000$ .”

The trap: Computing the right-hand limit by replacing $x$ with $3000$ in the formula on the LEFT. Each side has its own formula, and that is the whole point of the question.

Marking: 1 mark per one-sided limit, 1 mark for the equation set up, 1 mark for the value of the parameter, 1 mark for the interpretation.

Check before you hand in

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

The table of values around the point

On the calculator, display the function at four values that close in on the point, two on each side. The announced limit should show up in the decimals.

For $\lim_{x\to 3}\dfrac{x^{2}-9}{x-3}$ , the table at $2.99$ and $3.01$ shows $5.99$ and $6.01$ . A table showing $7.9$ would say the factoring is wrong.
The degrees, by eye

For a limit at infinity, keep only the dominant terms and compare. The result should come out in two seconds and confirm the detailed calculation.

$\dfrac{6x^{3}+x}{2x^{3}-5}$ reduces to $\dfrac{6x^{3}}{2x^{3}}=3$ . A detailed calculation that gave $0$ or infinity is wrong.
The two sides of an asymptote

Near a vertical asymptote, test a value just to the left and a value just to the right. The two signs are almost always opposite.

$\dfrac{1}{x-2}$ equals $-10{,}000$ at $1.9999$ and $+10{,}000$ at $2.0001$ . Two identical signs signal a squared factor in the denominator, or an error.
Plugging the parameter back in

Once the parameter is determined, recompute the two formulas at the junction. They must give exactly the same number.

With $k=900{,}000$ : $900-0.15\times 3000=450$ and $\dfrac{900{,}000}{2000}=450$ . Two different numbers signal a sign error in the equation.
The order of magnitude in context

Reread the answer in the units of the statement and ask whether it is plausible. An absurd economic limit almost always signals a degree error.

An average cost that tended to zero at infinity would say that producing no longer costs anything: it actually tends to the variable unit cost.

The typical problem, taken apart

A financial aid schedule made continuous, then studied in the long run

An aid program pays an amount $A(x)$ , in dollars, to a household whose annual income is $x$ , in dollars.

The schedule is $A(x)=900-0.15x$ for $0\le x\le 3000$ , and $A(x)=\dfrac{k}{x-1000}$ for $x>3000$ .

Determine $k$ so that the schedule is continuous at $3000$ , then study the amount paid to very high incomes.

Step 1

On $[0, 3000)$ and on $(3000, +\infty)$ , $A$ is continuous: the first formula is linear, the second is rational with a nonzero denominator. Only $x=3000$ is in question.

Why

Saying first where continuity poses NO problem, and why, is worth a mark and saves you from checking three conditions at points where they are automatic.

Step 2

The inequality $0\le x\le 3000$ is non-strict, so $A(3000)$ is computed with the first formula: $A(3000)=900-0.15\times 3000=900-450=450$ dollars.

Why

This is the first of the three conditions, and it is the formula that INCLUDES the point that gives it. Taking the other one is the classic junction error.

Step 3

Left-hand limit: $\lim_{x\to 3000^{-}}(900-0.15x)=450$ , equal to $A(3000)$ .

Why

The left formula is continuous, so its limit is its value. Saying it explicitly avoids giving the impression that you guessed.

Step 4

Right-hand limit: $\lim_{x\to 3000^{+}}\dfrac{k}{x-1000}=\dfrac{k}{2000}$ , because $3000-1000=2000$ and the denominator does not vanish.

Why

Each side is computed with ITS formula. This is the only step where the parameter appears, and it is therefore the one that carries the equation.

Step 5

Continuity requires the two one-sided limits to be equal: $\dfrac{k}{2000}=450$ , so $k=900{,}000$ .

Why

A single equation is enough here because the value $A(3000)$ already coincides with the left-hand limit. With two parameters, a system would have been needed.

Step 6

Check: $\dfrac{900{,}000}{3000-1000}=\dfrac{900{,}000}{2000}=450$ , which is indeed $A(3000)$ . The three conditions are satisfied.

Why

Plugging the value back in takes ten seconds and catches every sign error. It is also the sentence that closes the question on the marking scheme.

Step 7

In the long run: $\lim_{x\to+\infty}\dfrac{900{,}000}{x-1000}=0$ , since the degree of the top, $0$ , is lower than that of the bottom, $1$ . The line $y=0$ is a horizontal asymptote.

Why

The long-term question is read from the degrees, without calculation. It then expects a sentence in words, not just a number.

Step 8

Interpretation: the aid decreases toward zero without ever reaching it. A very well-off household receives an arbitrarily small, but strictly positive, amount.

Why

This is the interpretation point of the schedule, and it rests on the distinction of the chapter: tending to zero is not equalling zero.

The conclusion, written out

“The schedule is continuous at $3000$ for $k=900{,}000$ . In the long run, $\lim_{x\to+\infty}A(x)=0$ : the aid fades out gradually without ever becoming zero.”

The classic mistake on this problem: Computing the right-hand limit by replacing $x$ with $3000$ in $900-0.15x$ . You then find $450=450$ , a true identity that does not determine $k$ , and the paper announces that the schedule is continuous for every value of $k$ .

Learn by heart

• The limit exists if and only if both one-sided limits exist AND are EQUAL.
• Continuity at $a$ : $f(a)$ exists, the limit exists, the two are equal. Three conditions, in this order.
• $\dfrac{0}{0}$ is not a number: it is the order to factor or to use the conjugate.
• Limit at infinity of a quotient: degree of the top lower gives $0$ , equal degrees give the ratio of the leading coefficients, degree of the top higher gives infinity.
• Vertical asymptote: denominator toward $0$ and numerator NOT toward $0$ . The sign decides between the two infinities.
• A horizontal asymptote can be crossed. A vertical asymptote never.
• IVT: continuity on a CLOSED interval, $k$ between the two images, conclusion “at least one $c$ ”. Never uniqueness without monotonicity.
• L'Hospital's rule is not in the program of 201-103.

Frequently asked questions

How do you resolve a zero over zero indeterminate form?

You transform the expression, you do not compute. With polynomials, factor top and bottom by the factor that vanishes, cancel it, then substitute again. With a square root, multiply the numerator and the denominator by the conjugate expression. In both cases, the zero over zero form guarantees that a common factor exists.

How can you tell a vertical asymptote from a hole?

Look at the numerator at the same place. If the denominator vanishes and the numerator does not, it is a vertical asymptote. If both vanish, the common factor cancels and only a hole remains, that is, one missing point on an otherwise ordinary curve.

Can a curve cross its horizontal asymptote?

Yes, as many times as it likes. A horizontal asymptote describes what the function does very far out, not a barrier. A curve can oscillate around its limiting value, cutting it at each oscillation, then settle onto it. Only a vertical asymptote is a boundary, because the function is not defined there.

When do you use the Intermediate Value Theorem?

As soon as the statement asks you to show that a value is reached or that an equation has a solution, without asking which one. You need a function that is continuous on a closed interval, and the target value must lie between the images of the two endpoints. The conclusion announces at least one solution, never a single one.

How do you find the parameter that makes a schedule continuous?

Compute the left-hand limit with the left formula and the right-hand limit with the right formula, then write that the two are equal. This equality gives an equation whose unknown is the parameter. Then check that the value found does give back the value of the function at the junction point.

What is the difference between absolute gap and relative gap?

The absolute gap is a subtraction, in dollars or units. The relative gap is a ratio, in percent. The two can go in opposite directions: two budgets whose difference grows indefinitely can become almost identical in proportion. The statement always specifies which of the two it is asking for.

Practise it

Corrected exercises: Limits and continuity, 201-103

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.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: limits and continuity (201-103)

The essentials

Three different questions at a single point

The three conditions for continuity at aaa

Asymptotes, and what they really forbid

The rules in table form

Which limit, according to the form obtained after substitution

The mistakes that cost marks

1. Confusing the limit and the value of the function

2. Declaring a vertical asymptote as soon as the denominator vanishes

3. Treating 00\dfrac{0}{0}00​ as a result

4. Taking the horizontal asymptote for a barrier

5. Concluding uniqueness with the Intermediate Value Theorem

6. Applying the IVT without checking its hypotheses

7. Making a schedule continuous with a single equality

8. Confusing the absolute gap and the relative gap

9. Forgetting that the SIGN of the denominator decides between the two infinities

Which method to choose

Which technique, according to the form obtained after substitution

Which theorem, according to the verb in the statement

How the answer is expected to be written

Writing up an application of the Intermediate Value Theorem

Making a piecewise schedule continuous

Check before you hand in

The typical problem, taken apart

A financial aid schedule made continuous, then studied in the long run

Learn by heart

Frequently asked questions

How do you resolve a zero over zero indeterminate form?

How can you tell a vertical asymptote from a hole?

Can a curve cross its horizontal asymptote?

When do you use the Intermediate Value Theorem?

How do you find the parameter that makes a schedule continuous?

What is the difference between absolute gap and relative gap?

Corrected exercises: Limits and continuity, 201-103

See also

Looking for a 201-103 tutor in Montreal?

The three conditions for continuity at $a$

3. Treating $\dfrac{0}{0}$ as a result