Asymptotes

A small workshop bottles mineral water. Each month it pays $2000$ thousand đồng in fixed costs (rent, electricity), and each bottle costs another $5$ thousand đồng. If it makes $x$ bottles, the average cost per bottle is $A(x)=\frac{5x+2000}{x}\ \ (\text{thousand đồng}).$ The more bottles it makes, the lower the average cost. Can it ever drop below $5$ thousand đồng?

Horizontal asymptotes

Explore: when $x$ is very large

We compute $A(x)$ for a few large values of $x$ :

$x$	$100$	$1000$	$10\,000$	$100\,000$
$A(x)$	$25$	$7$	$5.2$	$5.02$

The values get closer and closer to $5$ but stay above $5$ , because $A(x)=5+\dfrac{2000}{x}$ . On the graph, as $x$ grows, the curve gets closer and closer to the line $y=5$ without touching it. Such a line is called an asymptote.

So to find horizontal asymptotes, we compute the two limits of $f$ at $+\infty$ and at $-\infty$ (see limits of functions in grade 11).

Vertical asymptotes

For $y=\dfrac{1}{x-2}$ , when $x$ is close to $2$ the denominator is very small, so $y$ is very large in absolute value: $y(2.01)=100$ , $y(1.99)=-100$ . The graph "shoots" up or down along the line $x=2$ .

For a rational function, we find the values that make the denominator zero, then check the limit at each of them: there is a vertical asymptote only when the limit is infinite.

Example Linear over linear

Find the asymptotes of the graph of $y=\dfrac{2x-1}{x+1}$ .

Solution. We compute the limits at infinity by dividing the numerator and the denominator by $x$ :

$\lim_{x\to\pm\infty}\frac{2x-1}{x+1}=\lim_{x\to\pm\infty}\frac{2-\frac1x}{1+\frac1x}=2.$

The denominator is zero at $x=-1$ , and the numerator there is $-3\neq 0$ . As $x\to-1^+$ , the denominator is positive and very small:

$\lim_{x\to-1^+}\frac{2x-1}{x+1}=-\infty,\qquad \lim_{x\to-1^-}\frac{2x-1}{x+1}=+\infty.$

So the graph has the horizontal asymptote $y=2$ and the vertical asymptote $x=-1$ .

The graph of

y=\dfrac{2x-1}{x+1}

and its two asymptotes

In general, the graph of $y=\dfrac{ax+b}{cx+d}$ (with $c\neq0$ , $ad-bc\neq0$ ) has the vertical asymptote $x=-\dfrac{d}{c}$ and the horizontal asymptote $y=\dfrac{a}{c}$ . The next example shows why we must check the limit and not just look at the denominator.

Example A zero denominator but no asymptote

Find the asymptotes of the graph of $y=\dfrac{x+1}{x^2-1}$ .

Solution. The denominator is zero at $x=1$ and $x=-1$ . For $x\neq\pm1$ , we can simplify:

$y=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}.$

At these two values, the limits give different results:

$\lim_{x\to1^+}\frac{1}{x-1}=+\infty,\qquad \lim_{x\to-1}\frac{1}{x-1}=-\frac12\ (\text{finite}).$

At infinity: $\lim\limits_{x\to\pm\infty}\dfrac{1}{x-1}=0$ .

So the graph has exactly two asymptotes: the vertical asymptote $x=1$ and the horizontal asymptote $y=0$ ; the line $x=-1$ is not an asymptote.

Oblique asymptotes

Some graphs do not approach a horizontal line but a slanted line as $x\to\pm\infty$ .

For $y=\dfrac{ax^2+bx+c}{mx+n}$ , the quickest way is polynomial division, writing $y=px+q+\dfrac{r}{mx+n}$ . The part $\dfrac{r}{mx+n}$ tends to $0$ as $x\to\pm\infty$ , so the oblique asymptote is $y=px+q$ .

The graph of

y=\dfrac{x^2-x+2}{x-1}

with the oblique asymptote

y=x

When division is awkward, we use the limit formulas from the definition. We try them on another function to see that both methods agree.

Example Using the limit formulas

Find the oblique asymptote of the graph of $y=\dfrac{2x^2+3x-1}{x+2}$ .

Solution. We compute $a$ and then $b$ as $x\to+\infty$ (as $x\to-\infty$ the result is exactly the same):

$a=\lim_{x\to+\infty}\frac{2x^2+3x-1}{x(x+2)}=2,\qquad b=\lim_{x\to+\infty}\left(\frac{2x^2+3x-1}{x+2}-2x\right)=\lim_{x\to+\infty}\frac{-x-1}{x+2}=-1.$

Check. Division gives $2x^2+3x-1=(x+2)(2x-1)+1$ , that is $y=2x-1+\dfrac{1}{x+2}$ .

So the graph has the oblique asymptote $y=2x-1$ .

A graph can have two different horizontal asymptotes on its two sides, as in the next example.

Example Two horizontal asymptotes

Find the horizontal asymptotes of the graph of $y=\dfrac{3x}{\sqrt{x^2+4}}$ .

Solution. Take $x^2$ out of the square root: $\sqrt{x^2+4}=|x|\sqrt{1+\frac{4}{x^2}}$ . The sign of $|x|$ differs on the two sides:

$\lim_{x\to+\infty}\frac{3x}{x\sqrt{1+\frac4{x^2}}}=3,\qquad \lim_{x\to-\infty}\frac{3x}{-x\sqrt{1+\frac4{x^2}}}=-3.$

The function is defined and continuous on $\mathbb{R}$ , so there is no vertical asymptote.

So the graph has two horizontal asymptotes: $y=3$ (as $x\to+\infty$ ) and $y=-3$ (as $x\to-\infty$ ).

The graph of

y=\dfrac{3x}{\sqrt{x^2+4}}

What asymptotes mean in real life

We return to the bottling question from the start.

Watch out Common mistakes

A zero denominator always gives a vertical asymptote. Wrong: " $y=\dfrac{x+1}{x^2-1}$ has two vertical asymptotes $x=\pm1$ ". Why it is wrong: at $x=-1$ the limit is $-\dfrac12$ , which is finite. Right: only $x=1$ .
Writing a vertical asymptote as $y=\dots$ . Wrong: the vertical asymptote is $y=-1$ . Why it is wrong: a vertical line has the form $x=x_0$ . Right: vertical asymptote $x=-1$ , horizontal asymptote $y=2$ .
Losing the remainder or a sign in the division. Wrong: " $y=\dfrac{x^2-3x+4}{x-2}$ has the oblique asymptote $y=x+1$ ". Why it is wrong: $x^2-3x+4=(x-2)(x-1)+2$ . Right: the oblique asymptote is $y=x-1$ .
Ignoring the sign of $|x|$ . Wrong: " $\sqrt{x^2+4}\approx x$ , so $y=\dfrac{3x}{\sqrt{x^2+4}}$ only has the horizontal asymptote $y=3$ ". Why it is wrong: when $x<0$ , $\sqrt{x^2}=-x$ . Right: there is also $y=-3$ as $x\to-\infty$ .

Key terms

English	Tiếng Việt
asymptote	đường tiệm cận
vertical asymptote	tiệm cận đứng
horizontal asymptote	tiệm cận ngang
oblique (slant) asymptote	tiệm cận xiên
one-sided limit	giới hạn một bên
limit at infinity	giới hạn tại vô cực
polynomial division	phép chia đa thức
average cost	chi phí trung bình

Summary

Horizontal asymptote $y=y_0$ : $\lim\limits_{x\to+\infty}f=y_0$ or $\lim\limits_{x\to-\infty}f=y_0$ . Compute both sides.
Vertical asymptote $x=x_0$ : a one-sided limit at $x_0$ is $\pm\infty$ . For a rational function: find the roots of the denominator, then check the limit.
Oblique asymptote $y=ax+b$ : $a=\lim\dfrac{f(x)}{x}$ , $b=\lim[f(x)-ax]$ ; for $\dfrac{ax^2+bx+c}{mx+n}$ , divide and drop the remainder.
$y=\dfrac{ax+b}{cx+d}$ : vertical asymptote $x=-\dfrac dc$ , horizontal asymptote $y=\dfrac ac$ .
In a figure, asymptotes are dashed lines; the graph gets close to them but need not cross or touch them.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Find the asymptotes of the graph of $y=\dfrac{3x+2}{x-4}$ .

What you need and how to start

What you need: $y=\dfrac{ax+b}{cx+d}$ has the vertical asymptote $x=-\dfrac dc$ and the horizontal asymptote $y=\dfrac ac$ .

Steps:

Compute the limits as $x\to\pm\infty$ .
Find the root of the denominator and check that the numerator is not $0$ there.

Show solution

$\lim\limits_{x\to\pm\infty}y=3$ . The denominator is $0$ at $x=4$ , where the numerator is $14\neq0$ , so $\lim\limits_{x\to4^+}y=+\infty$ .

Answer. Vertical asymptote $x=4$ , horizontal asymptote $y=3$ .

Exercise 2

Find the asymptotes of the graph of $y=\dfrac{1-2x}{x+3}$ .

What you need and how to start

What you need: the coefficient of $x$ in the numerator is $-2$ ; watch the order of the terms.

Steps:

Write the numerator as $-2x+1$ to read off $a=-2$ .
Find the root of the denominator.

Show solution

$\lim\limits_{x\to\pm\infty}y=\dfrac{-2}{1}=-2$ . At $x=-3$ the numerator is $7\neq0$ , so $\lim\limits_{x\to-3^+}y=+\infty$ .

Answer. Vertical asymptote $x=-3$ , horizontal asymptote $y=-2$ .

Exercise 3

Find the asymptotes of the graph of $y=\dfrac{x^2+2x+3}{x+1}$ .

What you need and how to start

What you need: polynomial division: if $y=px+q+\dfrac{r}{x+1}$ , the oblique asymptote is $y=px+q$ .

Steps:

Divide $x^2+2x+3$ by $x+1$ .
Find the vertical asymptote from the denominator.

Show solution

$x^2+2x+3=(x+1)(x+1)+2$ , so $y=x+1+\dfrac{2}{x+1}$ . The part $\dfrac{2}{x+1}\to0$ as $x\to\pm\infty$ . At $x=-1$ the numerator is $2\neq0$ .

Answer. Vertical asymptote $x=-1$ , oblique asymptote $y=x+1$ .

Exercise 4

How many asymptotes does the graph of $y=\dfrac{x^2-4}{x^2-3x+2}$ have? Which lines are they?

What you need and how to start

What you need: if a root of the denominator is also a root of the numerator, the fraction can be simplified, and there may be no vertical asymptote there.

Steps:

Factorise the numerator and the denominator, and simplify.
Check the limit at each root of the denominator and at infinity.

Show solution

$y=\dfrac{(x-2)(x+2)}{(x-1)(x-2)}=\dfrac{x+2}{x-1}$ for $x\neq2$ . At $x=2$ : $\lim\limits_{x\to2}y=4$ , which is finite. At $x=1$ : $\lim\limits_{x\to1^+}y=+\infty$ . At infinity: $\lim\limits_{x\to\pm\infty}y=1$ .

Answer. Two asymptotes: the vertical asymptote $x=1$ and the horizontal asymptote $y=1$ .

Exercise 5

Consider $y=\dfrac{x^2-3x+4}{x-2}$ . Is each claim true or false?

a) The line $x=2$ is a vertical asymptote of the graph.

b) The graph has no horizontal asymptote.

c) The line $y=x+1$ is an oblique asymptote of the graph.

d) The graph has exactly two asymptotes.

What you need and how to start

What you need: divide to write $y=px+q+\dfrac{r}{x-2}$ .

Steps:

Divide $x^2-3x+4$ by $x-2$ .
Check each claim.

Show solution

$x^2-3x+4=(x-2)(x-1)+2$ , so $y=x-1+\dfrac{2}{x-2}$ .

a) True: the numerator is $2\neq0$ at $x=2$ , so $\lim\limits_{x\to2^+}y=+\infty$ .

b) True: $y\to+\infty$ as $x\to+\infty$ and $y\to-\infty$ as $x\to-\infty$ .

c) False: the oblique asymptote is $y=x-1$ .

d) True: the vertical asymptote $x=2$ and the oblique asymptote $y=x-1$ .

Answer. a) True; b) True; c) False; d) True.

Exercise 6

Find the total number of vertical and horizontal asymptotes of the graph of $y=\dfrac{2x+1}{x^2-4}$ .

What you need and how to start

What you need: if the degree of the numerator is less than the degree of the denominator, the limit at infinity is $0$ .

Steps:

Find the roots of the denominator and check the numerator there.
Compute the limits at $\pm\infty$ .

Show solution

The denominator is $0$ at $x=\pm2$ ; the numerator there is $5$ and $-3$ , both non-zero, so there are two vertical asymptotes $x=2$ , $x=-2$ . $\lim\limits_{x\to\pm\infty}y=0$ , so there is the horizontal asymptote $y=0$ .

Answer. $3$ .

Exercise 7

A tank holds $200$ litres of clean water. Salt water with a concentration of $30$ grams per litre is pumped in at $10$ litres per minute. After $t$ minutes, the salt concentration in the tank is $C(t)=\dfrac{30t}{20+t}$ (grams per litre). Find the horizontal asymptote of the graph of $C(t)$ and explain what it means.

What you need and how to start

What you need: the amount of salt is $30\cdot10t=300t$ grams and the volume is $200+10t$ litres; the horizontal asymptote is the limit as $t\to+\infty$ .

Steps:

Check the formula: $\dfrac{300t}{200+10t}=\dfrac{30t}{20+t}$ .
Compute $\lim\limits_{t\to+\infty}C(t)$ .

Show solution

$\dfrac{300t}{200+10t}=\dfrac{30t}{20+t}$ (divide the numerator and the denominator by $10$ ). $\lim\limits_{t\to+\infty}\dfrac{30t}{20+t}=30$ . For example $C(10)=10$ , $C(60)=22.5$ , $C(180)=27$ .

Answer. Horizontal asymptote $C=30$ : over time the salt concentration in the tank gets close to $30$ grams per litre (the concentration of the incoming water) but never exceeds it.

Exercise 8 Review

Compute $\lim\limits_{x\to+\infty}\left(\sqrt{x^2+2x}-x\right)$ , and deduce a horizontal asymptote of the graph of $y=\sqrt{x^2+2x}-x$ .

What you need and how to start

What you need: multiply and divide by the conjugate $\sqrt{x^2+2x}+x$ to remove the form $\infty-\infty$ (limits of functions, grade 11).

Steps:

Write $\sqrt{x^2+2x}-x=\dfrac{2x}{\sqrt{x^2+2x}+x}$ .
Divide the numerator and the denominator by $x$ (with $x>0$ ).

Show solution

$\sqrt{x^2+2x}-x=\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}=\dfrac{2x}{\sqrt{x^2+2x}+x}=\dfrac{2}{\sqrt{1+\frac2x}+1}\to\dfrac{2}{2}=1$ .

Answer. The limit is $1$ ; the line $y=1$ is a horizontal asymptote (as $x\to+\infty$ ).

Exercise 9 Review

Prove that the graph of $y=\dfrac{2x^2-1}{x^2-2x+3}$ has no vertical asymptote, then find its horizontal asymptote.

What you need and how to start

What you need: a quadratic expression with $\Delta<0$ has no roots (see the sign of a quadratic expression).

Steps:

Compute the discriminant of the denominator.
Divide the numerator and the denominator by $x^2$ to find the limit at infinity.

Show solution

The denominator $x^2-2x+3$ has $\Delta=4-12=-8<0$ , so it is always positive: the function is defined and continuous on $\mathbb{R}$ and has no vertical asymptote. $\lim\limits_{x\to\pm\infty}y=\dfrac{2}{1}=2$ .

Answer. No vertical asymptote; horizontal asymptote $y=2$ .

Exercise 10 Challenge

Find the oblique asymptotes of the graph of $y=\sqrt{x^2+4x}$ .

What you need and how to start

What you need: $a=\lim\dfrac{f(x)}{x}$ , $b=\lim[f(x)-ax]$ , computed separately as $x\to+\infty$ and as $x\to-\infty$ ; when $x<0$ , $\sqrt{x^2}=-x$ .

Steps:

Find the domain.
Compute $a$ and $b$ as $x\to+\infty$ (use the conjugate).
Repeat as $x\to-\infty$ .

Show solution

The domain is $(-\infty, -4]\cup[0, +\infty)$ . As $x\to+\infty$ : $a=\lim\sqrt{1+\frac4x}=1$ , $b=\lim\left(\sqrt{x^2+4x}-x\right)=\lim\dfrac{4x}{\sqrt{x^2+4x}+x}=2$ . As $x\to-\infty$ : $\sqrt{x^2+4x}=-x\sqrt{1+\frac4x}$ , so $a=-1$ and $b=\lim\left(\sqrt{x^2+4x}+x\right)=\lim\dfrac{4x}{\sqrt{x^2+4x}-x}=\lim\dfrac{4}{-\sqrt{1+\frac4x}-1}=-2$ .

Answer. Two oblique asymptotes: $y=x+2$ (as $x\to+\infty$ ) and $y=-x-2$ (as $x\to-\infty$ ).

Horizontal asymptotes

Explore: when $x$ is very large

Vertical asymptotes

Oblique asymptotes

What asymptotes mean in real life

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The line $x=2$ is a vertical asymptote of the graph.	True	False
b) The line $y=2$ is a horizontal asymptote of the graph.	True	False
c) The line $y=2x+1$ is an oblique asymptote of the graph.	True	False
d) The oblique asymptote and the two coordinate axes form a triangle of area $\dfrac14$ .	True	False

Asymptotes

Horizontal asymptotes

Explore: when xxx is very large

Vertical asymptotes

Oblique asymptotes

What asymptotes mean in real life

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Explore: when $x$ is very large