MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: limits and the limit laws (MATH 140)

This sheet is not a summary of sections 2.2 and 2.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on limits in MATH 140 at McGill University, and which precise gesture avoids each loss.

Most limits of this chapter are computed correctly and then lose half their marks: for a missing condition $x \ne a$ , a law used without its permit, one side forgotten. Every value below is exact and done by hand, as on the exam, and every trap comes with the sentence that earns the method mark.

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The thread of the chapter

A limit describes what $f$ does NEAR $a$ , never AT $a$ : the value $f(a)$ is irrelevant, the form $\frac{0}{0}$ is an order to rewrite using $x \ne a$ , and every limit law needs its permit, the limits of the pieces must exist first.

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

Limit of a function
One-sided limits
Limit laws
Direct substitution property
Indeterminate form 0/0
Factor theorem
Conjugate of a radical expression
Absolute value function
Greatest integer function
Squeeze theorem
Fundamental trigonometric limit

The essentials

Three questions at one number, and the one that does not count

• At a number $a$ there are three separate questions: $\lim_{x\to a^-} f(x)$ , $\lim_{x\to a^+} f(x)$ , and the value $f(a)$ .
• $\lim_{x\to a} f(x) = L$ if and only if BOTH one-sided limits exist and are EQUAL to $L$ .
• The value $f(a)$ plays no part in the limit: $f$ may be undefined at $a$ , or defined with any other value.
• A corner does not stop a limit; a jump does. The limit reads heights, not directions.
• If $f(x) = g(x)$ for all $x \ne a$ near $a$ , then $f$ and $g$ have the same limit at $a$ . This is the licence to cancel a factor $x - a$ .

The curve approaches height

3

from both sides, so the limit at

2

3

; the red dot only says

f(2) = 1

, a different question.

On a graph, a full dot gives a value, an empty dot gives nothing but a height the curve approaches. Read the curve on each side, never the isolated dot.

The limit laws and their permits

• Sum, difference, constant multiple, product, power, root: the limit is computed piece by piece, IF each piece has a finite limit (and an even root needs a positive limit).
• Quotient: $\lim \frac{f}{g} = \frac{\lim f}{\lim g}$ only if $\lim g \ne 0$ .
• Direct substitution: for a polynomial, or a rational function with $a$ in its domain, the limit is the value $f(a)$ . It is the ONLY case where plugging in is a method.
• Backwards product law: if $q \to 0$ and $\frac{p}{q} \to L$ finite, then $p = \frac{p}{q} \cdot q \to 0$ . So a nonzero number over something tending to $0$ has no finite limit.
• The laws never reverse: $u + v$ or $uv$ can have a limit while $u$ and $v$ have none.

Name the law, or the direct substitution property, on the line where you use it: in MATH 140 the method marks are on those words.

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.

The quotient $\frac{f}{g}$ as $x \to a$ : when the law applies

Read a line as: with these limits for the numerator and the denominator, the limit of the quotient is the last column. The red lines are not answers: they are orders to rewrite the expression before any limit is taken.

$\lim f$	$\lim g$	$\lim \frac{f}{g}$
$L$	$M \ne 0$	$\frac{L}{M}$
Example: $\lim_{x\to -1} \frac{x^3 + 2x^2 - 5}{x^2 + 3} = \frac{-4}{4} = -1$ .
$L \ne 0$	$0$	no finite limit
Example: $\frac{x + 1}{x - 2}$ at $2$ : if it tended to $M$ , then $x + 1 = \frac{x+1}{x-2}(x - 2) \to M \cdot 0 = 0$ , not $3$ .
$0$	$0$	form $\frac{0}{0}$ settles nothing
Example: $\frac{x^3 - 8}{x^2 - 5x + 6} = \frac{x^2 + 2x + 4}{x - 3} \to -12$ at $2$ . Same form, other result: At $1$ , $\frac{x^2 - 1}{x - 1} \to 2$ while $\frac{(x - 1)^2}{x - 1} \to 0$ : same form, two answers. What to do: Factor (factor theorem), use the conjugate, or a common denominator; cancel $x - a$ using $x \ne a$ ; substitute.
no limit	no limit	law silent no law applies
Example: $u = \frac{x}{\|x\|}$ , $v = -\frac{x}{\|x\|}$ at $0$ : $u + v = 0 \to 0$ and $uv = -1 \to -1$ . What to do: Simplify the combination first, or squeeze it, then take the limit.

The second line is proved with the product law alone. Whether the quotient goes up or down without bound belongs to the next chapters of the course.

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. Giving the value of the function as its limit

the whole question, often 2 to 3 marks

What not to write

“ $f(-2) = 3$ , so $\lim_{x\to -2} f(x) = 3$ .”

What to write

“From both sides the curve approaches the height $1$ , so $\lim_{x\to -2} f(x) = 1$ ; the value $f(-2) = 3$ is not used.”

Why: The limit uses only $x \ne a$ . The point at $a$ can be moved, removed or doubled without the limit noticing, as the figure of the essentials shows.

2. Answering 1 to the form 0/0

the whole limit

What not to write

“Substitution gives $\frac{0}{0}$ , and a number over itself is $1$ , so the limit is $1$ .”

What to write

“The form is $\frac{0}{0}$ . For $x \ne 1$ , $\frac{x^2 - 1}{x - 1} = x + 1$ , so the limit is $2$ .”

Three quotients give

\frac{0}{0}

x = 1

: the blue one tends to

2

, the orange one to

0

, and the green one jumps from

-1

1

and has no limit.

Why: Numerator and denominator are two DIFFERENT quantities that both tend to $0$ . The form only says the quotient law has no permit; the answer comes from rewriting.

3. Cancelling without the condition x ≠ a

1 rigour mark, on every limit of this kind

What not to write

“ $\frac{x^2 - x - 6}{x - 3} = x + 2$ , so the limit at $3$ is $5$ .”

What to write

“For $x \ne 3$ , $\frac{x^2 - x - 6}{x - 3} = \frac{(x - 3)(x + 2)}{x - 3} = x + 2$ , so the limit at $3$ is $5$ .”

Why: The two functions differ at $3$ : one is undefined there. The condition $x \ne 3$ is exactly what makes the cancellation legal and the limit equal.

4. Dropping the absolute value of a square root

the whole question

What not to write

“ $\sqrt{x^2 - 4x + 4} = x - 2$ , so $\frac{\sqrt{x^2 - 4x + 4}}{x^2 - 4} = \frac{1}{x + 2} \to \frac{1}{4}$ .”

What to write

“ $\sqrt{(x - 2)^2} = |x - 2|$ . For $x > 2$ the quotient tends to $\frac{1}{4}$ , for $x < 2$ to $-\frac{1}{4}$ : the limit does not exist.”

Why: A square root is never negative, while $x - 2 < 0$ on the left of $2$ . Where the inside of $|u|$ vanishes, the two sides use two formulas.

5. Computing only one side of an absolute value

the whole question

What not to write

“ $\frac{|x + 4|}{(x + 1)(x + 4)} = \frac{1}{x + 1} \to -\frac{1}{3}$ as $x \to -4$ .”

What to write

“For $x > -4$ the quotient is $\frac{1}{x + 1} \to -\frac{1}{3}$ ; for $x < -4$ it is $-\frac{1}{x + 1} \to \frac{1}{3}$ . The sides differ: no limit.”

Why: $|x + 4| = x + 4$ only on the right of $-4$ . Split exactly where the inside vanishes, and nowhere else: $\frac{|2x - 1| - |2x + 1|}{x}$ needs no split at $0$ .

6. Using the product law on a factor with no limit

1 to 2 method marks, although the number is right

What not to write

“ $\lim_{x\to 0} x \sin\frac{1}{x} = \lim_{x\to 0} x \cdot \lim_{x\to 0} \sin\frac{1}{x} = 0$ .”

What to write

“ $\lim \sin\frac{1}{x}$ does not exist, so the product law does not apply. Since $-|x| \le x \sin\frac{1}{x} \le |x|$ and $\pm|x| \to 0$ , the squeeze theorem gives $0$ .”

The factor

\sin\frac{1}{x}

oscillates forever, yet the curve

y = x \sin\frac{1}{x}

is pinched between

y = |x|

and

y = -|x|

, which meet at the origin.

Why: The product law needs BOTH limits to exist. A bounded factor times a factor tending to $0$ is a squeeze, and the write-up must say so.

7. Squeezing between bounds that do not meet

the whole question

What not to write

“ $-1 \le \sin\frac{1}{x} \le 1$ , so by the squeeze theorem $\lim_{x\to 0} \sin\frac{1}{x}$ is between $-1$ and $1$ .”

What to write

“The bounds tend to $-1$ and $1$ , which differ, so the theorem says nothing. At $x = \frac{1}{n\pi}$ the value is $0$ , at $x = \frac{2}{(4n + 1)\pi}$ it is $1$ : no limit.”

Why: The squeeze theorem needs two bounds with the SAME limit, and it never asserts that a limit exists between two bounds.

8. Matching the sine with the wrong denominator

2 marks

What not to write

“ $\lim_{x\to 0} \frac{\sin 7x}{2x} = 1$ , since sine over something tends to $1$ .”

What to write

“ $\frac{\sin 7x}{2x} = \frac{7}{2} \cdot \frac{\sin 7x}{7x}$ , and $7x \to 0$ , so the limit is $\frac{7}{2}$ .”

Why: $\frac{\sin \theta}{\theta} \to 1$ needs the SAME $\theta$ in the sine and in the denominator, with $\theta \to 0$ , in radians. At $x \to \pi$ , $\frac{\sin x}{x - \pi} \to -1$ , not $1$ .

Which method to choose

Which gesture, by the FORM of the limit

Substitute a first, on the side. What you get picks the tool

If a number, and the function is a polynomial or a rational function defined at $a$ → done: direct substitution property, name it

Example: $\lim_{x\to -1} \frac{x^3 + 2x^2 - 5}{x^2 + 3} = -1$
If $\frac{0}{0}$ with polynomials → factor $x - a$ out of both (factor theorem, division), cancel for $x \ne a$

Example: $\frac{t^4 - 1}{t^3 - 1} \to \frac{4}{3}$ at $1$
If $\frac{0}{0}$ with a square root → multiply top AND bottom by the conjugate

Example: $\frac{\sqrt{x + 4} - 3}{x - 5} \to \frac{1}{6}$ at $5$
If $\frac{0}{0}$ with a cube root → use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Example: $\frac{\sqrt[3]{x} - 2}{x - 8} \to \frac{1}{12}$ at $8$
If a sum or difference of fractions, each with a denominator tending to 0 → common denominator first, then cancel

Example: $\frac{1}{x - 2} - \frac{4}{x^2 - 4} \to \frac{1}{4}$ at $2$
If $|u|$ , $\sqrt{u^2}$ or $\lfloor u \rfloor$ → check the sign of $u$ near $a$ ; split into one-sided limits only if $u$ vanishes (or is an integer) at $a$

Example: $\frac{|x + 4|}{x^2 + 5x + 4}$ : $\frac{1}{3}$ and $-\frac{1}{3}$ , no limit
If a bounded factor ( $\sin$ , $\cos$ of something wild, $\lfloor \cdot \rfloor$ ) times a factor tending to $0$ → squeeze, with two bounds that share their limit

Example: $x^2 \sin\frac{1}{x} \to 0$
If $\sin(\text{something})$ over something, the angle tending to $0$ → rewrite as a constant times $\frac{\sin \theta}{\theta}$ with the same $\theta$

Example: $\frac{\sin 3x}{\sin 5x} \to \frac{3}{5}$

After a cancellation, substitute AGAIN and read the new form: $\frac{x^3 - 3x + 2}{x^2 - 1}$ becomes $\frac{(x - 1)(x + 2)}{x + 1}$ , which gives $0$ , not a second $\frac{0}{0}$ .

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.

The squeeze theorem, written for full marks

When to use it: A factor with no limit ( $\sin\frac{1}{x}$ , $\cos\frac{\pi}{x}$ , $\lfloor \frac{1}{x} \rfloor$ ) multiplied by a factor tending to $0$ , or a function given only through inequalities

1 Start from the bound of the wild factor, valid for every $x \ne a$ : $-1 \le \sin\frac{1}{x} \le 1$ .
2 Multiply by the other factor, stating its sign; if it can be negative, use $|x|$ or treat the two sides separately and REVERSE the inequalities where it is negative.
3 Compute the limits of the two bounds, by the limit laws or direct substitution, and check that they are EQUAL.
4 Name the theorem and conclude.

Concluding sentence

“For all $x \ne 0$ , $-x^2 \le x^2 \sin\frac{1}{x} \le x^2$ . Since $\lim_{x\to 0} (-x^2) = \lim_{x\to 0} x^2 = 0$ , the squeeze theorem gives $\lim_{x\to 0} x^2 \sin\frac{1}{x} = 0$ .”

The trap: Writing $\lim x^2 \cdot \lim \sin\frac{1}{x}$ : the second limit does not exist, and the product law has no permit.

Marking: Typically 1 mark for the bounds, 1 for their common limit, 1 for naming the theorem; the number alone earns nothing.

Proving that a limit does not exist

When to use it: The one-sided formulas differ (absolute value, floor, piecewise), or the function oscillates

1 Say why the two sides must be treated separately: the inside of $|u|$ vanishes at $a$ , or the formula changes at $a$ .
2 Compute $\lim_{x\to a^-}$ with the formula valid for $x < a$ , and $\lim_{x\to a^+}$ with the one valid for $x > a$ .
3 Compare: two different numbers, or one side without a limit.
4 For an oscillation, exhibit two families of $x$ tending to $a$ on which the function takes two different values.

Concluding sentence

“ $\lim_{x\to -4^-} \frac{|x + 4|}{x^2 + 5x + 4} = \frac{1}{3}$ and $\lim_{x\to -4^+} \frac{|x + 4|}{x^2 + 5x + 4} = -\frac{1}{3}$ . The one-sided limits are different, so the limit does not exist.”

The trap: Writing “the limit does not exist because the function is not defined at $a$ ”: that is never a reason, since the limit ignores $a$ .

Marking: Usually 1 mark per one-sided limit and 1 for the conclusion with its reason.

Check before you hand in

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

Expand the factorization before cancelling

Multiply the factors back out, or test them at two simple values of x. One value can agree by accident.

$x^3 - 8 = (x - 2)^3$ agrees at $x = 0$ ( $-8$ both sides) but fails at $x = 1$ : $-7$ against $-1$ .
The sign on one side

Take x just to the right of a and read the sign of every factor. The limit cannot have the opposite sign.

$\frac{x - 4}{2 - \sqrt x}$ with $x$ slightly above $4$ : positive over negative, so $-4$ is plausible and $4$ is not.
A convenient nearby value, by hand

Choose x near a so that every root is exact, and compute the quotient. It will not prove the limit, but it catches a factor or a sign.

For $\frac{\sqrt{x + 3} - 2}{x - 1}$ , take $x = 0.61$ : $\frac{1.9 - 2}{-0.39} \approx 0.256$ , close to $\frac{1}{4}$ .
The squeeze bounds really meet

Substitute a into both bounds before writing the conclusion. If the two numbers differ, the theorem gives nothing.

$4x - 9$ and $x^2 - 4x + 7$ both give $7$ at $x = 4$ ; $-1$ and $1$ never meet.

The typical problem, taken apart

A limit at a junction: conjugate on one side, absolute value on the other

Let $f(x) = \frac{\sqrt{x + 3} - 2}{x - 1}$ for $-3 \le x < 1$ , $f(1) = 1$ , and $f(x) = \frac{x^2 - 1}{8|x - 1|}$ for $x > 1$ .

Find $\lim_{x\to 1} f(x)$ , if it exists, and compare it with $f(1)$ . No calculator.

Step 1

The formula changes at $1$ , so compute $\lim_{x\to 1^-} f(x)$ and $\lim_{x\to 1^+} f(x)$ separately. The value $f(1) = 1$ is set aside.

Why

The junction forces the split; the value at $1$ answers a different question, and using it now is the first trap of the chapter.

Step 2

Left: substitution gives $\frac{0}{0}$ . With the conjugate, $\frac{(x + 3) - 4}{(x - 1)\left(\sqrt{x + 3} + 2\right)} = \frac{1}{\sqrt{x + 3} + 2}$ for $x \ne 1$ , so $\lim_{x\to 1^-} f(x) = \frac{1}{2 + 2} = \frac{1}{4}$ .

Why

A root in a $\frac{0}{0}$ calls for the conjugate, on top AND bottom. The condition $x \ne 1$ licenses the cancellation.

Step 3

Right: for $x > 1$ , $x - 1 > 0$ so $|x - 1| = x - 1$ , and $f(x) = \frac{(x - 1)(x + 1)}{8(x - 1)} = \frac{x + 1}{8}$ . So $\lim_{x\to 1^+} f(x) = \frac{2}{8} = \frac{1}{4}$ .

Why

On this side only one case of the absolute value exists; saying which one, with the sign of $x - 1$ , is the justification.

Step 4

Both one-sided limits equal $\frac{1}{4}$ , so $\lim_{x\to 1} f(x) = \frac{1}{4}$ , while $f(1) = 1 \ne \frac{1}{4}$ .

Why

The limit exists because the sides AGREE, not because $f(1)$ is defined; it differs from the value, and nothing forbids that.

Step 5

Check with values chosen for exact roots: $x = 0.61$ gives $\frac{1.9 - 2}{-0.39} \approx 0.256$ ; $x = 1.2$ gives $\frac{1.44 - 1}{8(0.2)} = \frac{0.44}{1.6} = 0.275$ . Both are near $0.25$ .

Why

A quick check on each side catches a lost factor $8$ or a wrong sign in the absolute value, at no cost.

The conclusion, written out

“ $\lim_{x\to 1^-} f(x) = \frac{1}{4}$ and $\lim_{x\to 1^+} f(x) = \frac{1}{4}$ , so $\lim_{x\to 1} f(x) = \frac{1}{4}$ , although $f(1) = 1$ .”

The classic mistake on this problem: Answering $1$ because $f(1) = 1$ ; or writing $|x - 1| = -(x - 1)$ on the right and finding $-\frac{1}{4}$ , then concluding that the limit does not exist.

Learn by heart

• $\lim_{x\to a} f(x) = L$ iff both one-sided limits exist and EQUAL $L$ . The value $f(a)$ plays no part.
• Every limit law needs its permit: each piece has a finite limit, and for a quotient $\lim g \ne 0$ .
• Plugging in is a method only for polynomials and rational functions defined at $a$ : name the direct substitution property.
• $\frac{0}{0}$ is an order to rewrite: factor, conjugate, common denominator, then cancel for $x \ne a$ .
• $\sqrt{u^2} = |u|$ . Split into two sides only where the inside of $|u|$ vanishes.
• Squeeze: two bounds, SAME limit. Multiplying an inequality by a negative number reverses it.
• $\frac{\sin \theta}{\theta} \to 1$ as $\theta \to 0$ , in radians, with the same $\theta$ above and below.
• If $q \to 0$ and $\frac{p}{q}$ has a finite limit, then $p \to 0$ .

Frequently asked questions

Why is the limit not just the value of the function at that point?

Because a limit only looks at the values of the function for x close to a, never at a itself. The function may be undefined at a, or defined with a completely different value, and the limit does not change. The two coincide for polynomials and for rational functions defined at a, which is the direct substitution property, but not in general.

What do I do when I plug in and get 0 over 0?

Do not stop and do not answer 1. Zero over zero means the quotient law cannot be used yet. Rewrite the expression: factor the polynomials, multiply by the conjugate if there is a square root, or put fractions over a common denominator. Cancel the common factor, which is allowed because x is different from a, then substitute again.

How do I show that a limit does not exist?

Compute the limit from the left and the limit from the right separately, using the formula valid on each side. If they are two different numbers, the limit does not exist. For a function that oscillates, like the sine of one over x, find two sequences of x approaching a on which the function takes two different values.

When should I use the squeeze theorem instead of the limit laws?

Use it when one factor has no limit but stays bounded, such as the sine or cosine of something that blows up, and the other factor tends to zero. The product law cannot be used then. Bound the bounded factor between minus one and one, multiply by the other factor, and check that both bounds have the same limit before concluding.

Does the limit of sin x over x equal 1 in degrees?

No. The limit equals 1 only when x is measured in radians. In degrees, the sine of x degrees over x tends to pi over 180, because x degrees is pi x over 180 radians. In MATH 140 every angle is in radians unless the question says otherwise, and the rewriting must match the angle inside the sine with the denominator.

Practise it

Corrected exercises: Limits and the limit laws, 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 Exponential, inverse and logarithmic functions Next sheet Continuity and the Intermediate Value Theorem

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

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Revision sheet: limits and the limit laws (MATH 140)

The essentials

Three questions at one number, and the one that does not count

The limit laws and their permits

The rules in table form

The quotient fg\frac{f}{g}gf​ as x→ax \to ax→a: when the law applies

The mistakes that cost marks

1. Giving the value of the function as its limit

2. Answering 1 to the form 0/0

3. Cancelling without the condition x ≠ a

4. Dropping the absolute value of a square root

5. Computing only one side of an absolute value

6. Using the product law on a factor with no limit

7. Squeezing between bounds that do not meet

8. Matching the sine with the wrong denominator

Which method to choose

Which gesture, by the FORM of the limit

How the answer is expected to be written

The squeeze theorem, written for full marks

Proving that a limit does not exist

Check before you hand in

The typical problem, taken apart

A limit at a junction: conjugate on one side, absolute value on the other

Learn by heart

Frequently asked questions

Why is the limit not just the value of the function at that point?

What do I do when I plug in and get 0 over 0?

How do I show that a limit does not exist?

When should I use the squeeze theorem instead of the limit laws?

Does the limit of sin x over x equal 1 in degrees?

Corrected exercises: Limits and the limit laws, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?

The quotient $\frac{f}{g}$ as $x \to a$ : when the law applies