MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: limits involving infinity and asymptotes (MATH 203)

This sheet is not a summary of section 2.6 of Thomas' Calculus: you already have the textbook and the lecture notes. It answers one question, what makes students lose marks on limits involving infinity and asymptotes in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rules of the chapter fit on a few lines: divide by the dominant power, read the sign on each side of a vertical asymptote, divide to find an oblique asymptote. The marks go elsewhere, in the algebra that has to happen BEFORE a rule applies: a term left undivided, an exponent subtracted wrongly, $|x|$ forgotten, a denominator never factored. That is what the traps below are made of. No L'Hôpital's Rule anywhere: it comes much later in the course, and the algebra is faster.

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

The limit rules of this chapter are short; the marks are lost in the algebra before them: every term divided by the dominant power, $\sqrt{x^2} = |x|$ at $-\infty$ , the denominator factored before its signs are read.

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

Limit at infinity
Infinite limit
Horizontal asymptote
Vertical asymptote
Oblique asymptote
Dominant terms
Indeterminate form
Rational function
Conjugate expression
Long division of polynomials
Laws of exponents
Sandwich Theorem

The essentials

At infinity: every term divided, every exponent subtracted

• For a quotient as $x \to \pm\infty$ , divide EVERY term of the numerator and of the denominator by the highest power of $x$ in the DENOMINATOR, then let each $\frac{c}{x^r}$ ( $r > 0$ ) go to $0$ .
• Write each term as a power first: $\sqrt{x} = x^{1/2}$ , $\frac{1}{x^3} = x^{-3}$ , and divide with $\frac{x^a}{x^b} = x^{a - b}$ . At infinity, $x^{-1}$ is LARGER than $x^{-3}$ .
• Compare the largest exponent on top, $p$ , with the largest below, $q$ : $p < q$ gives $0$ ; $p = q$ gives the ratio of their coefficients; $p = q + 1$ gives an oblique asymptote; $p > q + 1$ gives $\pm\infty$ with no asymptote line.
• Exponentials: $e^x$ dominates at $+\infty$ but tends to $0$ at $-\infty$ ; divide by the exponential that dominates IN THAT DIRECTION, with $\frac{e^{a}}{e^{b}} = e^{a - b}$ .

The first line of the division, “dividing by $x^2$ , the highest power of the denominator”, is what the marker looks for; the answer alone earns little.

The square root of x squared, and the two directions

• $\sqrt{x^2} = |x|$ . Factoring $x^2$ out of a root gives $\sqrt{9x^2 + 4} = |x|\sqrt{9 + \frac{4}{x^2}}$ , which is $x\sqrt{\cdots}$ for $x > 0$ and $-x\sqrt{\cdots}$ for $x < 0$ .
• An odd root keeps the sign: $\sqrt[3]{x^3} = x$ for every $x$ , so $\sqrt[3]{8x^3 - x^2} = x\sqrt[3]{8 - \frac{1}{x}}$ at both ends.
• A root never splits over a sum: $\sqrt{x^2 + 5} \ne x + \sqrt 5$ . A difference $\sqrt{A} - \sqrt{B}$ of form $\infty - \infty$ goes through the conjugate.
• The two ends are TWO questions: a function has at most two horizontal asymptotes, and an oblique asymptote can exist in one direction only.

\sqrt{4x^2 + 1} - x

follows

y = x

on the right and

y = -3x

on the left:

\sqrt{4x^2} = 2|x|

2x

on one side and

-2x

on the other.

Before accepting a limit at $-\infty$ , look at the sign of the expression at $x = -1000$ : it catches the forgotten minus sign 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.

A quotient at infinity, read on its largest exponents

Read a line as: when the largest exponent on top and the largest below compare as in the first two columns, the limit at $+\infty$ is the third. The red line is not an answer yet: the sign still has to be decided.

Top exponent	Bottom exponent	Limit at infinity
$p$	$q > p$	$0$
Example: $\frac{3x + 1}{x^{3/2} + 2} \to 0$ , since $1 < \frac{3}{2}$ .
$p$	$q = p$	$\frac{a}{b}$
Example: $\frac{4x - x^{2/3} + 1}{3x + \sqrt{x}} \to \frac{4}{3}$ , and $\frac{6x^{-1} + x^{-3}}{2x^{-2} - x^{-1}} \to \frac{6}{-1} = -6$ (both largest exponents are $-1$ ).
$p$	$q = p - 1$	$\pm\infty$ , oblique line
Example: $\frac{2x^2 - 3x - 1}{x - 2} = 2x + 1 + \frac{1}{x - 2}$ : asymptote $y = 2x + 1$ .
$p$	$q < p - 1$	$\pm\infty$ , no line
Example: $\frac{x^4}{x^2 + 1}$ behaves like $x^2$ : $+\infty$ at both ends, a parabola, and no line fits, since $4 - 2 = 2$ .
$\sqrt{x^2 + \cdots}$	$x$ , at $-\infty$	sign of $\|x\|$ sign still to decide
Example: $\frac{\sqrt{9x^2 + 4}}{2x - 5} \to -\frac{3}{2}$ as $x \to -\infty$ . Same form, other result: The same quotient tends to $+\frac{3}{2}$ as $x \to +\infty$ : same exponents, opposite answers. What to do: Write $\sqrt{x^2 + \cdots} = -x\sqrt{1 + \cdots}$ for $x < 0$ before dividing.

The table gives the limit; the division still has to be written, because it is the division that earns the method marks.

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. Dividing only some of the terms

the whole limit, 2 to 3 marks

What not to write

“ $\lim_{x\to\infty} \frac{5x^2 + 3x}{2x^2 - 7} = \lim_{x\to\infty} \frac{5 + 3x}{2 - 7} = -\infty$ .”

What to write

“Dividing every term by $x^2$ : $\frac{5 + \frac{3}{x}}{2 - \frac{7}{x^2}} \to \frac{5}{2}$ .”

Why: A fraction bar is a pair of parentheses: $\frac{5x^2 + 3x}{x^2}$ is $\frac{5x^2}{x^2} + \frac{3x}{x^2}$ . Dividing the leading terms only changes the function.

2. Taking a negative exponent for the dominant one

the whole limit

What not to write

“In $\frac{3x^{-1} + x^{-2}}{x^{-1} - 4x^{-3}}$ the highest powers are $x^{-2}$ and $x^{-3}$ , so the limit is $\frac{1}{-4}$ .”

What to write

“Multiplying top and bottom by $x^3$ : $\frac{3x^2 + x}{x^2 - 4} \to 3$ .”

Why: At infinity $x^{-3} = \frac{1}{x^3}$ is the SMALLEST term. Rewriting the negative exponents as fractions, then clearing them, removes the doubt.

3. Writing the root of x squared as x at minus infinity

the whole limit, and the second asymptote

What not to write

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

What to write

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

Why: $\sqrt{x^2} = |x| = -x$ when $x$ is negative. At $x = -1000$ the root is positive and $3x + 1$ negative: $+\frac{1}{3}$ was impossible from the start.

4. Splitting a square root over a sum

2 marks

What not to write

“ $\sqrt{x^2 + 9} = x + 3$ , so $\lim_{x\to\infty} \left(\sqrt{x^2 + 9} - x\right) = 3$ .”

What to write

“ $\sqrt{x^2 + 9} - x = \frac{9}{\sqrt{x^2 + 9} + x} \to 0$ .”

Why: $\sqrt{a + b} \ne \sqrt a + \sqrt b$ : at $x = 4$ , $\sqrt{25} = 5$ , not $7$ . The conjugate turns the difference into a quotient whose denominator grows.

5. Declaring an asymptote where the factor cancels

1 to 2 marks, and a wrong sketch

What not to write

“The denominator of $\frac{x - 1}{x^2 - 1}$ vanishes at $1$ and $-1$ : two vertical asymptotes.”

What to write

“ $\frac{x - 1}{x^2 - 1} = \frac{1}{x + 1}$ for $x \ne 1$ : a hole at $\left(1, \frac{1}{2}\right)$ , and one vertical asymptote, $x = -1$ .”

\frac{x - 1}{x^2 - 1}

has only one branch going to infinity, at

x = -1

; at

x = 1

the graph just misses the point

\left(1, \frac{1}{2}\right)

Why: A vertical asymptote needs the numerator NOT to tend to $0$ . Factor the denominator ( $x^2 - 1 = (x - 1)(x + 1)$ ) and the numerator before deciding anything.

6. Misapplying an exponent law

the whole limit

What not to write

“ $\frac{e^{3x}}{e^x} = e^3$ , so $\lim_{x\to\infty} \frac{e^{3x}}{e^x + 4} = e^3$ .”

What to write

“ $\frac{e^{3x}}{e^x} = e^{3x - x} = e^{2x}$ ; dividing every term by $e^x$ , $\frac{e^{2x}}{1 + 4e^{-x}} \to \infty$ .”

Why: The law is $\frac{e^a}{e^b} = e^{a - b}$ : exponents are SUBTRACTED, and the variable stays. $e^3$ comes from dividing $3x$ by $x$ , which is not a law of exponents.

7. Believing ln x levels off

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

What not to write

“The graph of $\ln x$ is almost flat after $x = 50$ , so it has a horizontal asymptote near $y = 4$ .”

What to write

“For every $M$ , $\ln x > M$ as soon as $x > e^M$ , so $\lim_{x\to\infty} \ln x = \infty$ : no horizontal asymptote.”

The red points

(e, 1)

(e^2, 2)

(e^3, 3)

(e^4, 4)

: each new level costs a factor

e

x

, but every level is reached.

Why: Slow growth is not bounded growth. $\ln x$ passes $4$ at $e^4 \approx 54.6$ , $10$ at $e^{10} \approx 22\,026$ , and every level after that.

8. Dividing polynomials without placeholder or sign change

2 marks

What not to write

“ $x^3 + 1$ divided by $x^2 - x$ gives the quotient $x - 1$ , so the oblique asymptote is $y = x - 1$ .”

What to write

“ $x^3 + 0x^2 + 0x + 1 = (x^2 - x)(x + 1) + (x + 1)$ , so $\frac{x^3 + 1}{x^2 - x} = x + 1 + \frac{x + 1}{x^2 - x}$ : the oblique asymptote is $y = x + 1$ .”

Why: Subtracting $x(x^2 - x) = x^3 - x^2$ leaves $+x^2$ , not $-x^2$ : every sign of the subtracted line changes. Expanding $(x^2 - x)(x + 1) + x + 1$ back gives $x^3 + 1$ and catches the slip.

Which method to choose

Which tool, by the FORM of the limit

Look at where x goes and at what each piece of the formula does before writing anything: the form picks the tool

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

Example: $\frac{3 - 2x}{x^2 - 4}$ : $+\infty$ as $x \to 2^-$ , $-\infty$ as $x \to 2^+$
If $x \to a$ , numerator and denominator $\to 0$ → factor both, cancel, then decide: a finite limit is a hole

Example: $\frac{x^3 - 8}{x^2 - 4} = \frac{x^2 + 2x + 4}{x + 2} \to 3$ at $2$
If $x \to \pm\infty$ , powers of $x$ (integer, negative or fractional) → rewrite as powers, divide every term by the dominant power of the denominator

Example: $\frac{4x - x^{2/3} + 1}{3x + \sqrt{x}} \to \frac{4}{3}$
If $x \to -\infty$ with $\sqrt{x^2 + \cdots}$ → write $\sqrt{x^2 + \cdots} = -x\sqrt{1 + \cdots}$ first

Example: $\frac{\sqrt{9x^2 + 4}}{2x - 5} \to -\frac{3}{2}$
If a difference of roots, both $\to \infty$ → multiply and divide by the conjugate

Example: $\sqrt{x^2 + 3x} - \sqrt{x^2 - x} \to 2$ at $\infty$
If exponentials $e^{kx}$ → divide by the exponential that dominates in THAT direction

Example: $\frac{5 - e^x}{1 + e^x}$ : $-1$ at $\infty$ , $5$ at $-\infty$
If a bounded oscillating factor times something $\to 0$ → Sandwich Theorem

Example: $-e^{-x} \le e^{-x}\cos x \le e^{-x}$ , so the limit at $\infty$ is $0$
If $x \to \infty$ inside $\frac{1}{x}$ , form $\infty \cdot 0$ → substitute $t = \frac{1}{x}$ , $t \to 0^+$

Example: $3x\tan\frac{1}{x} = \frac{3\tan t}{t} \to 3$

L'Hôpital's Rule is section 4.5, at the end of the course: every limit of this chapter must go through one of these branches.

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 near the asymptotes”, or opens a curve sketch

1 Factor the numerator and the denominator completely, and state the domain.
2 At each excluded point $a$ : if a factor cancels completely, compute the limit of the simplified formula, a hole with its coordinates. Otherwise give the two one-sided limits with the sign of each factor: $x = a$ is a vertical asymptote.
3 Compute $\lim_{x\to\infty} f(x)$ and $\lim_{x\to -\infty} f(x)$ SEPARATELY, dividing every term by the dominant power; each finite value $L$ gives $y = L$ .
4 If the numerator's degree is one 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 $f(x) - (mx + b)$ .

Concluding sentence

“Since $\lim_{x\to -1^-} f(x) = +\infty$ and $\lim_{x\to -1^+} f(x) = -\infty$ , the line $x = -1$ is a vertical asymptote; since $f(x) - (x + 1) = -\frac{1}{x + 1} \to 0$ as $x \to \pm\infty$ , the line $y = x + 1$ is an oblique asymptote at both ends.”

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

Marking: Typically 1 mark per asymptote with its justification, 1 for the side of the curve, and the method marks in the factoring and the division.

Check before you hand in

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

The calculator at a large x

MATH 203 allows a scientific calculator: evaluate the expression at x = 1000 and at x = -1000. It proves nothing, but a value far from your limit, or of the wrong sign, says the algebra is wrong.

$\frac{\sqrt{9x^2 + 4}}{2x - 5}$ at $x = -1000$ gives about $-1.496$ : the limit $-\frac{3}{2}$ is plausible, and $+\frac{3}{2}$ is ruled out.
A test value on each side of a vertical asymptote

Evaluate the expression 0.001 to the left and 0.001 to the right of the asymptote and read only the sign. It confirms each branch in ten seconds.

$\frac{3 - 2x}{x^2 - 4}$ at $x = 1.999$ is about $+250$ : $+\infty$ on the left, and a written $-\infty$ would be a sign error.
Expand a division back

After a long division, multiply the quotient by the divisor and add the remainder: you must get the numerator back, term by term.

$(x - 2)(2x + 1) + 1 = 2x^2 - 3x - 2 + 1 = 2x^2 - 3x - 1$ : the division of $2x^2 - 3x - 1$ by $x - 2$ is right.
Plug in one number after a simplification

Every rewriting (factoring, cancelling, splitting a fraction) must give the same value at one simple point, say x = 2.

$\frac{x^3 - 8}{x^2 - 4}$ and $\frac{x^2 + 2x + 4}{x + 2}$ at $x = 3$ : $\frac{19}{5}$ both times. A factorization $(x - 2)(x^2 + 4)$ would give $\frac{13}{5}$ .

The typical problem, taken apart

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

Find all the asymptotes of $f(x) = \frac{x^3 + x^2 - 2x}{x^2 - 1}$ , with the one-sided limits, and say where the graph has a hole. Give the position of the graph relative to its oblique asymptote.

Every step justified as on a MATH 203 final; the calculator is only for checking.

Step 1

Factor: $x^3 + x^2 - 2x = x(x^2 + x - 2) = x(x + 2)(x - 1)$ and $x^2 - 1 = (x - 1)(x + 1)$ . Domain: $x \ne 1$ , $x \ne -1$ . For $x \ne 1$ , $f(x) = \frac{x(x + 2)}{x + 1}$ .

Why

Factoring first shows which zeros of the denominator are shared with the numerator: the candidates for a hole. Taking out the common factor $x$ before factoring the trinomial is the step most often skipped.

Step 2

At $x = 1$ both the numerator and the denominator vanish. After cancelling, $\lim_{x\to 1} f(x) = \frac{1 \cdot 3}{2} = \frac{3}{2}$ : a hole at $\left(1, \frac{3}{2}\right)$ , no asymptote.

Why

A finite limit at an excluded point is a hole, and naming it with its coordinates earns its own mark.

Step 3

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

Why

The NEGATIVE numerator reverses both signs; checking only the denominator gives the two answers backwards.

Step 4

Divide: $x^2 + 2x = (x + 1)(x + 1) - 1$ , so $f(x) = x + 1 - \frac{1}{x + 1}$ for $x \ne 1$ . Since $\frac{1}{x + 1} \to 0$ at both ends, $y = x + 1$ is an oblique asymptote at both ends, and there is no horizontal one.

Why

The quotient of the division IS the asymptote; reading $\frac{x^3}{x^2} = x$ off the leading terms gives the slope and misses the $+1$ .

Step 5

Position: $f(x) - (x + 1) = -\frac{1}{x + 1}$ , negative for $x > -1$ and positive for $x < -1$ : below the asymptote on the right, above it on the left. Check at $x = 3$ : $f(3) = \frac{27 + 9 - 6}{8} = \frac{15}{4}$ and $3 + 1 - \frac{1}{4} = \frac{15}{4}$ .

Why

The sign of the remainder makes the sketch right near the asymptote, and the check at one point catches a division error.

The conclusion, written out

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

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

Learn by heart

• At $\pm\infty$ : divide EVERY term by the highest power of the denominator; $\frac{x^a}{x^b} = x^{a - b}$ .
• $\sqrt{x^2} = |x|$ : at $-\infty$ , $\sqrt{x^2 + \cdots} = -x\sqrt{1 + \cdots}$ ; $\sqrt[3]{x^3} = x$ always.
• $\frac{L}{0}$ with $L \ne 0$ : infinite, with the sign of each factor on EACH side; factor the denominator first.
• Numerator and denominator both $0$ at $a$ : factor, cancel; a finite limit is a hole, not an asymptote.
• Numerator degree one more: long division with $0$ placeholders; the quotient is the oblique asymptote.
• $\infty - \infty$ , $\frac{\infty}{\infty}$ , $0 \cdot \infty$ are forms, not values; $\ln x \to \infty$ and has no horizontal asymptote.

Frequently asked questions

How do I find a limit at infinity of a rational function in MATH 203?

Divide every term of the numerator and of the denominator by the highest power of x that appears in the denominator, then let each term of the form a constant over a power of x go to zero. If the top degree is smaller the limit is zero, if the degrees are equal it is the ratio of the leading coefficients, and if the top degree is larger the limit is infinite.

Why is the square root of x squared equal to minus x at 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 goes to minus infinity, the root becomes minus x times the root of what is left. A cube root does not do this: the cube root of x cubed is x for every x.

What do I do with negative or fractional exponents in a limit at infinity?

Write every term as a power of x first. A negative exponent is a fraction, so x to the minus three is one over x cubed, which is the smallest term at infinity, not the largest. Then either multiply the numerator and the denominator to clear the fractions, or divide every term by the dominant power, subtracting exponents.

When does a function have an oblique asymptote?

A rational function has one exactly when the degree of the numerator is one more than the degree of the denominator. Divide the polynomials, writing a zero for every missing power: the quotient, a line mx plus b, is the oblique asymptote, and the sign of the remainder over the denominator tells you if the curve is above or below it.

Can I use L'Hopital's Rule for these limits in MATH 203?

Not in this chapter. L'Hopital's Rule is taught near the end of the course, and the questions on limits at infinity and asymptotes are graded on the algebraic methods: dividing by the dominant term, the conjugate, factoring, combining logarithms. They are also faster than the rule on rational functions and square roots.

Practise it

Corrected exercises: Limits involving infinity and asymptotes, 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 One-sided limits Next sheet Continuity

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

Looking for a MATH 203 tutor in Montreal?

Get in touch for a first session. Asymptotes come back in every curve sketching question of the final, and the algebra that decides them is exactly what a few sessions fix.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: limits involving infinity and asymptotes (MATH 203)

The essentials

At infinity: every term divided, every exponent subtracted

The square root of x squared, and the two directions

The rules in table form

A quotient at infinity, read on its largest exponents

The mistakes that cost marks

1. Dividing only some of the terms

2. Taking a negative exponent for the dominant one

3. Writing the root of x squared as x at minus infinity

4. Splitting a square root over a sum

5. Declaring an asymptote where the factor cancels

6. Misapplying an exponent law

7. Believing ln x levels off

8. Dividing polynomials without placeholder or sign change

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 a limit at infinity of a rational function in MATH 203?

Why is the square root of x squared equal to minus x at minus infinity?

What do I do with negative or fractional exponents in a limit at infinity?

When does a function have an oblique asymptote?

Can I use L'Hopital's Rule for these limits in MATH 203?

Corrected exercises: Limits involving infinity and asymptotes, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?