MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: infinite limits, limits at infinity and asymptotes (MATH 140)

This sheet is not a summary of sections 2.2 and 2.6 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on infinite limits, limits at infinity and asymptotes in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter comes before the derivative, so every limit is settled by algebra: a sign study, a division by the dominant term, a conjugate, a long division. The algebra is rarely what goes wrong. What goes wrong is a sign that was never studied, a direction that was never distinguished, and a form such as $\infty - \infty$ treated as a number. Every value below is exact and done by hand, as on the exam.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

Infinity is not a number you substitute: near a vertical asymptote the SIGN of the denominator on each side decides, and at infinity the DOMINANT term decides, with a sign that depends on the direction, since $\sqrt{x^2} = |x|$ .

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

Infinite limit
Vertical asymptote
Limit at infinity
Horizontal asymptote
Oblique asymptote
One-sided limits
Indeterminate form
Rational function
Conjugate expression
Long division of polynomials
Sign chart
Removable discontinuity

The essentials

A zero of the denominator: three possible stories

• If $N(x) \to L \ne 0$ and $D(x) \to 0$ as $x \to a$ , then $\left|\frac{N}{D}\right| \to \infty$ , and the SIGN on each side is the sign of $\frac{L}{D(x)}$ : study $D$ on the left and on the right of $a$ .
• A factor $(x - a)$ to an ODD power changes sign at $a$ : the two branches go in opposite directions. To an EVEN power it does not: both branches go the same way.
• If $N(a) = D(a) = 0$ , the form is $\frac{0}{0}$ : factor, cancel, then decide. A factor that cancels completely leaves a HOLE, not an asymptote.
• $x = a$ is a vertical asymptote as soon as ONE of the one-sided limits is $\infty$ or $-\infty$ : $e^{1/x}$ has one at $0$ from the right only.
• $\lim_{x\to a} f(x) = \infty$ is written only when BOTH sides go to $+\infty$ , and even then the limit does not exist: the notation says how it fails.

The denominator of

\frac{x-1}{(x-1)(x+2)(x-2)^2}

vanishes three times: opposite branches at

x = -2

, both branches up at

x = 2

, and only a hole at

x = 1

The marker reads the sign study, not the final symbol: two lines, “ $x - 3 \to 0^-$ on the left, $0^+$ on the right”, carry most of the marks of an infinite limit.

At infinity: the dominant term, with its sign

• Rational function: divide by the highest power of the DENOMINATOR. Degree of the numerator below: limit $0$ ; equal: ratio of the leading coefficients; exactly one more: oblique asymptote by long division; more than that: no asymptote line.
• $\sqrt{x^2} = |x|$ . For $x > 0$ , $\sqrt{x^2 + \dots} = x\sqrt{1 + \dots}$ ; for $x < 0$ , $\sqrt{x^2 + \dots} = -x\sqrt{1 + \dots}$ .
• $e^x \to \infty$ at $+\infty$ and $e^x \to 0$ at $-\infty$ ; the dominant exponential is the largest exponent at $+\infty$ and the most negative one at $-\infty$ .
• $\ln x \to \infty$ at $\infty$ and $\to -\infty$ at $0^+$ ; $\arctan x \to \frac{\pi}{2}$ at $\infty$ and $-\frac{\pi}{2}$ at $-\infty$ .
• The two ends are TWO questions: a function has at most two horizontal asymptotes, one per direction, and a graph may cross them at finite $x$ .

Before any algebra, decide the sign of the answer from the formula at a large negative $x$ : it catches the missing minus of $\sqrt{x^2}$ every time.

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.

Forms you meet in this chapter, and which ones are answers

Read a line as: when the limit takes the form of the first column, in the situation of the second, the result is the third. A red cell is not an answer: it is an order to transform the expression, or to study a sign.

Form	Where	Result
$\frac{L}{\pm\infty}$	$L$ finite	$0$
Example: $\lim_{x\to\infty} \frac{5}{x^2 + 1} = 0$ and $\lim_{x\to\infty} \frac{4x - 1}{x^3 + 2} = 0$ .
$\frac{L}{0}$	$L \ne 0$ , $x \to a$	$\pm\infty$ sign still to decide
Example: $\frac{2x}{x - 3}$ : $-\infty$ as $x \to 3^-$ , $+\infty$ as $x \to 3^+$ , and $\frac{x - 5}{(x + 1)^2} \to -\infty$ on both sides of $-1$ . What to do: Write the sign of the denominator on each side ( $0^-$ or $0^+$ ), then the sign of the quotient.
$\frac{0}{0}$	$x \to a$	form $\frac{0}{0}$ settles nothing
Example: $\frac{x^2 - 4}{x^2 - x - 2} = \frac{x + 2}{x + 1} \to \frac{4}{3}$ at $2$ : a hole. Same form, other result: $\frac{(x - 2)^2}{x - 2} \to 0$ while $\frac{x - 2}{(x - 2)^2}$ has a vertical asymptote at $2$ : same form, opposite outcomes. What to do: Factor $(x - a)$ in the numerator and the denominator, cancel, and look again.
$\frac{\infty}{\infty}$	$x \to \pm\infty$	form $\frac{\infty}{\infty}$ settles nothing
Example: $\frac{3x^2 - 5x + 1}{2x^2 + 7} \to \frac{3}{2}$ after dividing by $x^2$ . Same form, other result: $\frac{2x}{x} \to 2$ while $\frac{x^2}{x} \to \infty$ and $\frac{x}{x^2} \to 0$ . What to do: Divide the numerator and the denominator by the dominant term of the denominator.
$\infty - \infty$	$x \to \pm\infty$	form $\infty - \infty$ settles nothing
Example: $\sqrt{x^2 + 6x} - x \to 3$ as $x \to \infty$ , by the conjugate. Same form, other result: $\sqrt{x^2 + 6x} - x \to 3$ while $\sqrt{x^2 + 6} - x \to 0$ and $x^2 - x \to \infty$ . What to do: Factor out the dominant term, or multiply by the conjugate, or combine two logarithms into one.
$-\infty - \infty$	both terms $\to -\infty$	$-\infty$
Example: $x^3 - 100x^2 \to -\infty$ as $x \to -\infty$ : both terms push the same way, nothing to decide.
$L \cdot \infty$	$L \ne 0$	$\pm\infty$
Example: $x^3\left(1 - \frac{100}{x}\right) \to +\infty$ as $x \to \infty$ , since the bracket tends to $1 > 0$ .

A limit of the form $\frac{L}{0}$ is not indeterminate, since its size is certain, but it is not an answer either until the sign is written: that is why it is red.

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. Writing nonzero over zero equals infinity, with no side

2 marks, the whole limit

What not to write

“ $\lim_{x\to 3} \frac{2x}{x - 3} = \frac{6}{0} = \infty$ .”

What to write

“As $x \to 3$ , $2x \to 6 > 0$ ; $x - 3 \to 0^-$ on the left and $0^+$ on the right, so the limit is $-\infty$ from the left and $+\infty$ from the right, and $\lim_{x\to 3} \frac{2x}{x-3}$ does not exist.”

Why: $\frac{6}{0}$ is not a number and $\infty$ is not its value. The size of the quotient is certain, its sign is not, and a two-sided $= \infty$ asserts a sign on both sides that is false here.

2. Declaring an asymptote where the factor cancels

2 marks, and a wrong sketch

What not to write

“The denominator of $\frac{x^2 - 4}{x^2 - x - 2}$ vanishes at $x = 2$ , so $x = 2$ is a vertical asymptote.”

What to write

“ $\frac{x^2 - 4}{x^2 - x - 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)} = \frac{x + 2}{x + 1}$ for $x \ne 2$ , which tends to $\frac{4}{3}$ : a hole at $\left(2, \frac{4}{3}\right)$ , no asymptote. At $x = -1$ the numerator tends to $1 \ne 0$ : vertical asymptote.”

Why: A vertical asymptote needs the numerator NOT to tend to $0$ . When both vanish, factor first: the essentials figure shows the hole at $x = 1$ next to two genuine asymptotes.

3. Forgetting the minus sign of the square root of x squared

the whole limit, and the second horizontal asymptote

What not to write

“ $\lim_{x\to -\infty} \frac{4x + 3}{\sqrt{x^2 + 1}} = \lim_{x\to -\infty} \frac{4 + \frac{3}{x}}{\sqrt{1 + \frac{1}{x^2}}} = 4$ .”

What to write

“For $x < 0$ , $\sqrt{x^2 + 1} = |x|\sqrt{1 + \frac{1}{x^2}} = -x\sqrt{1 + \frac{1}{x^2}}$ , so the quotient is $\frac{4 + \frac{3}{x}}{-\sqrt{1 + \frac{1}{x^2}}} \to -4$ .”

\frac{4x+3}{\sqrt{x^2+1}}

settles on

y = 4

to the right and on

y = -4

to the left: two directions, two asymptotes.

Why: Bringing $x$ under the root as $x^2$ is only valid for $x > 0$ . For a large negative $x$ the numerator is negative and the root positive, so $+4$ was impossible from the start.

4. Answering 0 to infinity minus infinity

2 marks

What not to write

“For large $x$ , $\sqrt{x^2 + 5x} \approx x$ , so $\lim_{x\to\infty} \left(\sqrt{x^2 + 5x} - x\right) = 0$ .”

What to write

“ $\sqrt{x^2 + 5x} - x = \frac{5x}{\sqrt{x^2 + 5x} + x} = \frac{5}{\sqrt{1 + \frac{5}{x}} + 1} \to \frac{5}{2}$ .”

Why: $\sqrt{x^2 + 5x} \approx x$ is true as a ratio, but a difference of two huge numbers is decided by exactly what the approximation throws away. The conjugate turns the difference into a quotient that can be divided.

5. Believing a graph never crosses its horizontal asymptote

1 to 2 marks, on a true or false or a sketch

What not to write

“The curve of $\frac{\sin x}{x}$ crosses the line $y = 0$ , so $y = 0$ cannot be its horizontal asymptote.”

What to write

“ $\left|\frac{\sin x}{x}\right| \le \frac{1}{x}$ for $x > 0$ , so $\frac{\sin x}{x} \to 0$ and $y = 0$ is a horizontal asymptote, although the graph crosses it at every $x = k\pi$ .”

\frac{\sin x}{x}

is squeezed toward

0

and still meets the line

y = 0

at every red point

x = k\pi

Why: A horizontal asymptote is a statement about $x \to \pm\infty$ only. At finite $x$ the graph may cross it any number of times; even a rational function can, like $\frac{3x^2 - 5x + 1}{2x^2 + 7}$ at $x = -\frac{19}{10}$ .

6. Computing one end and assuming the other

1 mark per missing asymptote

What not to write

“ $\lim_{x\to\infty} \arctan x = \frac{\pi}{2}$ , so the horizontal asymptote is $y = \frac{\pi}{2}$ .”

What to write

“ $\arctan x \to \frac{\pi}{2}$ as $x \to \infty$ and $\arctan x \to -\frac{\pi}{2}$ as $x \to -\infty$ : the horizontal asymptotes are $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ .”

Why: For a rational function the two ends agree, and that habit is what fails on roots, exponentials and arctangent. Always write two limits, one per direction.

7. Dividing by the wrong exponential

the whole limit

What not to write

“ $\lim_{x\to -\infty} \frac{2e^x + 3}{e^x - 1} = \lim_{x\to -\infty} \frac{2 + 3e^{-x}}{1 - e^{-x}} = 2$ .”

What to write

“As $x \to -\infty$ , $e^x \to 0$ , so $\frac{2e^x + 3}{e^x - 1} \to \frac{0 + 3}{0 - 1} = -3$ .”

Why: At $-\infty$ , $e^x$ is not dominant, it vanishes; dividing by it makes $e^{-x}$ explode in both terms and the form $\frac{\infty}{\infty}$ reappears. The answer $2$ comes from dividing a second time and treating $\frac{\infty}{\infty}$ as a number.

8. Reading an oblique asymptote off the leading terms

2 marks

What not to write

“ $\frac{x^2 + x + 2}{x - 1} \approx \frac{x^2}{x} = x$ , so the oblique asymptote is $y = x$ .”

What to write

“ $x^2 + x + 2 = (x - 1)(x + 2) + 4$ , so $f(x) = x + 2 + \frac{4}{x - 1}$ and $f(x) - (x + 2) \to 0$ : the oblique asymptote is $y = x + 2$ .”

Why: The leading terms give the slope $m$ only; the intercept $b$ comes from the next term, and only the division, or the limit of $f(x) - mx$ , finds it. Here $f(x) - x = \frac{2x + 2}{x - 1} \to 2$ , not $0$ .

Which method to choose

Which tool, by the FORM of the limit

Look at where x goes and at what the pieces of the formula do, before writing anything: the form picks the tool

If $x \to a$ , numerator $\to L \ne 0$ , denominator $\to 0$ → sign of the denominator on each side, then the sign of the quotient

Example: $\frac{x - 5}{(x + 1)^2} \to -\infty$ on both sides of $-1$
If $x \to a$ , numerator and denominator $\to 0$ → factor $(x - a)$ , cancel, then decide: a finite limit is a hole

Example: $\frac{x^2 - 4}{x^2 - x - 2} \to \frac{4}{3}$ at $2$
If $x \to \pm\infty$ , a rational function → divide by the highest power of the denominator; one degree more on top means long division

Example: $\frac{2x^3 - x}{5 - x^2}$ behaves like $-2x$ : $-\infty$ at $\infty$ , $+\infty$ at $-\infty$
If $x \to -\infty$ with $\sqrt{x^2 + \dots}$ → write $\sqrt{x^2 + \dots} = -x\sqrt{1 + \dots}$ before dividing

Example: $\frac{6x - 1}{\sqrt{4x^2 + 1}} \to -3$
If a difference $\sqrt{A} - B$ with both terms $\to \infty$ → multiply and divide by the conjugate

Example: $\sqrt{4x^2 + x} - 2x \to \frac{1}{4}$
If exponentials $e^{kx}$ → divide by the dominant exponential FOR THAT DIRECTION

Example: $\frac{e^{3x} - e^{-x}}{e^{3x} + e^x}$ : $1$ at $\infty$ , $-\infty$ at $-\infty$
If a difference of two logarithms, both $\to \infty$ → combine into $\ln\frac{A}{B}$ , then use the continuity of $\ln$

Example: $\ln(2x + 1) - \ln(x + 3) \to \ln 2$
If $\arctan$ , $e$ or $\ln$ of an expression → limit of the inside first, then the outside function at that limit

Example: $\arctan\left(\frac{1}{x}\right)$ : $-\frac{\pi}{2}$ from the left, $\frac{\pi}{2}$ from the right

L'Hospital's rule belongs to chapter 4 of the course: every limit of this chapter must be done by one of these branches, and they are faster anyway.

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.

Finding all the asymptotes of a function

When to use it: Any question that says “find all the asymptotes”, “describe the behaviour of $f$ near its asymptotes”, or opens a sketch

1 State the domain, and list the points where the formula is undefined.
2 At each such point $a$ , compute the limit of the numerator. If it is nonzero, study the sign of the denominator on each side and give the two one-sided limits: $x = a$ is a vertical asymptote. If it is zero, factor and cancel: a finite limit gives a hole.
3 Compute $\lim_{x\to\infty} f(x)$ and $\lim_{x\to -\infty} f(x)$ SEPARATELY. Each finite value $L$ gives a horizontal asymptote $y = L$ .
4 If a limit is infinite and the numerator of a rational function has one degree more, divide: the quotient $mx + b$ is the oblique asymptote, since the remainder over the denominator tends to $0$ .
5 Give the side: the sign of $f(x) - L$ or of $f(x) - (mx + b)$ says whether the curve is above or below.

Concluding sentence

“Since $\lim_{x\to 2^-} f(x) = -\infty$ and $\lim_{x\to 2^+} f(x) = +\infty$ , the line $x = 2$ is a vertical asymptote; since $f(x) - (x + 3) = \frac{6}{x - 2} \to 0$ as $x \to \pm\infty$ , the line $y = x + 3$ is an oblique asymptote at both ends.”

The trap: Stopping at “the denominator vanishes at $1$ and $2$ ” without checking the numerator: one of them may be a hole.

Marking: Typically 1 mark per asymptote found, 1 per sign study justified, and 1 for the position of the curve; the method marks are in the sign studies.

Check before you hand in

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

A test value on each side

Plug a number just left and just right of the asymptote into the factors, mentally, and read only the signs. It confirms the direction of each branch in five seconds.

$\frac{2x}{x - 3}$ at $x = 2.9$ : $\frac{5.8}{-0.1} < 0$ , so $-\infty$ on the left; if you had written $+\infty$ , the sign study is wrong.
The sign at a large negative x

Before accepting a limit at minus infinity, look at the sign of the expression for x = -1000. A positive answer for a negative expression is a forgotten minus of the square root of x squared.

$\frac{6x - 1}{\sqrt{4x^2 + 1}}$ at $x = -1000$ : negative numerator, positive root, so the limit must be negative: $-3$ , not $3$ .
The numerator at a zero of the denominator

Evaluate the numerator at every zero of the denominator. Zero means factor and look for a hole; nonzero confirms the asymptote.

$x^2 - 4$ at $x = 2$ gives $0$ : hole. At $x = -1$ it gives $-3 \ne 0$ : asymptote.
One large value for an infinity minus infinity

Compute the expression once at x = 100, keeping only what you can do by hand. It will not prove the limit, but it catches an answer of 0 that should be 3.

$\sqrt{10600} - 100 \approx 2.96$ since $103^2 = 10609$ : close to $3$ , far from $0$ .

The typical problem, taken apart

A full census: a hole, a vertical asymptote and an oblique one

Find all the asymptotes of $f(x) = \frac{x^3 - x}{x^2 - 3x + 2}$ , with the one-sided limits, and say where the graph has a hole. Give the position of the curve relative to its oblique asymptote.

No calculator, no L'Hospital's rule: every step justified as on a MATH 140 final.

Step 1

Factor: $x^3 - x = x(x - 1)(x + 1)$ and $x^2 - 3x + 2 = (x - 1)(x - 2)$ . The domain is $x \ne 1$ , $x \ne 2$ , and for $x \ne 1$ , $f(x) = \frac{x(x + 1)}{x - 2}$ .

Why

Factoring first shows at once which zeros of the denominator are shared with the numerator: those are the candidates for a hole, the others for an asymptote.

Step 2

At $x = 1$ , the numerator and the denominator of the original formula both vanish. After cancelling, $\lim_{x\to 1} f(x) = \frac{1 \cdot 2}{1 - 2} = -2$ : the graph has a hole at $(1, -2)$ and no asymptote there.

Why

This is the step most students skip. A finite limit at an excluded point is a hole, and saying so, with its coordinates, is worth a mark on its own.

Step 3

At $x = 2$ , the numerator $x(x + 1) \to 6 > 0$ and $x - 2 \to 0^-$ on the left, $0^+$ on the right: $\lim_{x\to 2^-} f(x) = -\infty$ , $\lim_{x\to 2^+} f(x) = +\infty$ . The line $x = 2$ is a vertical asymptote.

Why

The two signs are the justification; “ $\frac{6}{0} = \infty$ ” would lose them. The factor $(x - 2)$ has an odd power, so the branches go opposite ways.

Step 4

At $\pm\infty$ the degree on top is one more, so divide: $x^2 + x = (x - 2)(x + 3) + 6$ , hence $f(x) = x + 3 + \frac{6}{x - 2}$ for $x \ne 1$ . Since $\frac{6}{x - 2} \to 0$ at both ends, $y = x + 3$ is an oblique asymptote at both ends, and there is no horizontal asymptote.

Why

The quotient of the division IS the asymptote. Reading $\frac{x^2}{x} = x$ off the leading terms would have given the slope and missed the $+3$ .

Step 5

Position: $f(x) - (x + 3) = \frac{6}{x - 2}$ is positive for $x > 2$ and negative for $x < 2$ : the curve is above its asymptote on the right, below on the left. Check at $x = 3$ : $f(3) = \frac{12}{1} = 12$ and $3 + 3 + 6 = 12$ .

Why

The sign of the remainder is what makes a sketch correct near the asymptote, and the check at one point catches a division error before it propagates.

The conclusion, written out

“The graph of $f$ has a hole at $(1, -2)$ , the vertical asymptote $x = 2$ with $\lim_{x\to 2^-} f(x) = -\infty$ and $\lim_{x\to 2^+} f(x) = +\infty$ , and the oblique asymptote $y = x + 3$ at both ends, the curve lying below it for $x < 2$ and above it for $x > 2$ .”

The classic mistake on this problem: Declaring two vertical asymptotes, $x = 1$ and $x = 2$ , because the denominator vanishes twice; or giving $y = x$ as the oblique asymptote from the leading terms.

Learn by heart

• $\frac{L}{0}$ with $L \ne 0$ : infinite, and the sign of the denominator on EACH side decides.
• Numerator and denominator both $0$ at $a$ : factor first; a finite limit is a hole, not an asymptote.
• One infinite one-sided limit is enough for a vertical asymptote.
• Rational function at $\pm\infty$ : divide by the highest power of the denominator; one degree more on top means an oblique asymptote by long division.
• $\sqrt{x^2} = |x|$ : at $-\infty$ , $\sqrt{x^2 + \dots} = -x\sqrt{1 + \dots}$ .
• $\infty - \infty$ and $\frac{\infty}{\infty}$ are forms, not values: conjugate, factor, divide, combine logarithms.
• Two ends, two limits: at most two horizontal asymptotes, and a graph may cross them.

Frequently asked questions

How do I find the vertical asymptotes of a rational function?

Factor the numerator and the denominator completely. Every zero of the denominator that is not cancelled by the numerator gives a vertical asymptote; a zero that cancels completely gives a hole instead. At each asymptote, find the sign of the denominator just to the left and just to the right to decide whether each branch goes up or down.

Why does the square root of x squared become minus x when x goes to minus infinity?

Because the square root of x squared is the absolute value of x, and for a negative x the absolute value is minus x. So when you factor x squared out of a square root and x is negative, the root becomes minus x times the root of what is left. Forgetting this minus sign is the classic way to find the wrong horizontal asymptote on the left.

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes what the graph does as x goes to plus or minus infinity, so the graph may cross the line at finite values of x, even infinitely often. The function sine of x over x tends to zero and still crosses the x-axis at every multiple of pi. A vertical asymptote is different: the function goes to infinity there.

When does a rational function have an oblique asymptote?

Exactly when the degree of the numerator is one more than the degree of the denominator. Do the long division: the quotient, a linear expression mx plus b, is the oblique asymptote, and the remainder over the denominator tends to zero. Its sign tells you whether the curve lies above or below the asymptote on each side.

Can I use L'Hospital's rule for limits at infinity in MATH 140?

L'Hospital's rule comes later in the course, with the applications of the derivative. For this chapter, the expected methods are algebraic: divide by the dominant term, multiply by the conjugate, factor, or combine logarithms. They are faster than the rule on rational functions and roots, and they are what the solutions to these questions are graded on.

Practise it

Corrected exercises: Limits at infinity and asymptotes, 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 Continuity and the Intermediate Value Theorem Next sheet The derivative as a limit

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

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Revision sheet: infinite limits, limits at infinity and asymptotes (MATH 140)

The essentials

A zero of the denominator: three possible stories

At infinity: the dominant term, with its sign

The rules in table form

Forms you meet in this chapter, and which ones are answers

The mistakes that cost marks

1. Writing nonzero over zero equals infinity, with no side

2. Declaring an asymptote where the factor cancels

3. Forgetting the minus sign of the square root of x squared

4. Answering 0 to infinity minus infinity

5. Believing a graph never crosses its horizontal asymptote

6. Computing one end and assuming the other

7. Dividing by the wrong exponential

8. Reading an oblique asymptote off the leading terms

Which method to choose

Which tool, by the FORM of the limit

How the answer is expected to be written

Finding all the asymptotes of a function

Check before you hand in

The typical problem, taken apart

A full census: a hole, a vertical asymptote and an oblique one

Learn by heart

Frequently asked questions

How do I find the vertical asymptotes of a rational function?

Why does the square root of x squared become minus x when x goes to minus infinity?

Can a graph cross its horizontal asymptote?

When does a rational function have an oblique asymptote?

Can I use L'Hospital's rule for limits at infinity in MATH 140?

Corrected exercises: Limits at infinity and asymptotes, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?