A limit at infinity asks what a function settles toward as x runs far out. If it approaches a fixed value L, the line y = L is a horizontal asymptote. Where the denominator is zero and the numerator is not, the values run away: a vertical asymptote.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Try it first
Work out f(x) = (3x + 5) ÷ (x + 1) at x = 10, at x = 100 and at x = 1000. Which description fits the three values?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Why the leading terms decide
At x = 1000, (2x + 1)/(x − 3) = 2001/997 ≈ 2.007. The +1 and −3 are swamped. Divide top and bottom by x: (2 + 1/x)/(1 − 3/x), and the fractions vanish as x → ∞, leaving 2.
The degree rule
Equal degrees: the ratio of leading coefficients. Smaller top: 0. Larger top: no horizontal asymptote, the function grows without bound. 3x²/(x³ + 1) → 0; x³/(x + 1) → ∞.
They may be crossed
A horizontal asymptote describes long-run behavior only. A curve may cross it near the origin and still approach it far out. Nothing forbids the crossing.
Vertical asymptotes
Set the denominator to zero. If the numerator is not zero there, the values run away and the limit does not exist. Check each side: (2x + 1)/(x − 3) → +∞ from the right of 3 and −∞ from the left.
Asymptote or hole
If top and bottom share the zero, the factor cancels and the graph has a hole rather than an asymptote. (x² − 9)/(x − 3) has a hole at 3, not an asymptote. Factor before deciding.
Which way it runs
Just right of 3, x − 3 is a small positive and the top is near 7: a large positive. Just left, a small negative: a large negative. Test a value on each side.
Step 2: Try It Yourself
Tap and try it out.
Find every asymptote and hole of f(x) = (x² − 4)/(x² − x − 2)
- 1(x − 2)(x + 2)/((x − 2)(x + 1))factor top and bottom
Step 0 of 4
- Point(3, 0.67)
Step 3: In Real Life
A drug in the bloodstream
After a dose, concentration rises then decays toward zero: a horizontal asymptote. A dose model with a vertical asymptote would mean an infinite level, so pharmacologists check the far ends of the graph first.
Step 4: Watch an Example
One step at a time.
Watch Ravi Read Both Ends of f(x) = (x² − 1) ÷ (x² − 4x + 3)
A fresh function with three things to say: one break that is a hole, one that is a genuine asymptote, and a far end that settles. Ravi factors before he decides anything, and finishes by checking every claim with a value.
- Step 1
Ravi factors both halves before deciding anything, because a zero of the bottom means two completely different things depending on what the top does there. The top is a difference of two squares, (x − 1)(x + 1). The bottom, x² − 4x + 3, needs two numbers multiplying to 3 and adding to −4, which are −1 and −3: (x − 1)(x − 3). The bottom is zero at x = 1 and at x = 3, and the two halves plainly share the factor (x − 1).
Watch Maya Read f(x) = (2x + 8) ÷ (x² − 16)
The same three questions on a function whose two halves have different degrees, so the far end behaves differently. Maya factors and cancels, and leaves you the three lines that finish the page.
- Step 1
Maya factors first, out of habit and for good reason. The top, 2x + 8, is 2(x + 4). The bottom is a difference of two squares, (x − 4)(x + 4). The bottom is zero at x = 4 and at x = −4, so there are two places where the original rule says nothing, and the factored form shows at a glance that only one of them is shared with the top.
Watch Ana and Leo Take Two Roads to the Same Far End
The far end of f(x) = (6x² − x) ÷ (3x² + 5). Ana divides top and bottom by a power of x; Leo compares degrees. The two roads agree, and the last question is what happens when they are handed a function whose two degrees do not match, where the one-line ratio has nothing to say.
- Step 1
Ana takes the longer road. Dividing top and bottom by x², the highest power in sight, changes nothing about the value of the fraction — it is one division applied to both halves, which is multiplying by 1 in disguise. The top becomes 6 − 1/x and the bottom becomes 3 + 5/x².
Step 5: Your Turn
Practice makes it stick.
The Chair Workshop
Problem 1 of 2
A workshop pays $900 a week for its rent and tools, plus $12 of wood for every chair it makes, so a week in which it makes x chairs costs 12x + 900 dollars. The average cost of a chair is that total shared among the x chairs. As the workshop makes more and more chairs, what does the average cost of a chair settle toward, in dollars?
dollars per chair
The Park Clean-Up
Problem 2 of 2
A council is quoted C(p) = 6000p ÷ (100 − p) dollars to clear p percent of the litter from a park. It has $54,000 to spend. What is the largest whole percent it can afford to clear?
percent
Ends and Breaks
For a far end, the highest power on each half decides. For a break, the zeros of the bottom decide — after the top has been checked at each of them.
1 of 5
What does (7x + 2) ÷ (x − 5) settle toward as x grows without bound?
2 of 5
What does (4x + 9) ÷ (x² + 3) settle toward as x grows without bound?
3 of 5
f(x) = (x + 5) ÷ (x² − 25). At which x is its vertical asymptote?
4 of 5
What does (5x³ + 2x) ÷ (2x³ − x + 1) settle toward as x grows without bound? Type it as a decimal.
5 of 5
f(x) = (x² + 1) ÷ (x − 2). What does its far end do?
Whose Step?
One line on each page is the first to go wrong. Find it, then put it right.
1 of 6
Maya listed the asymptotes of f(x) = (x² − 9) ÷ (x² − 3x). Which line is the first to go wrong?
Maya’s asymptotes of f(x) = (x² − 9)/(x² − 3x)
- 1x² − 3x = 0 at x = 0 and x = 3
- 2vertical asymptotes at x = 0 and x = 3
- 3degrees equal, leading coefficients 1 and 1
- Answerasymptotes: x = 0, x = 3, y = 1
2 of 6
At which x does f(x) = (x² − 9) ÷ (x² − 3x) have its one genuine vertical asymptote?
3 of 6
Dev found the horizontal asymptote of f(x) = (3x² + 5) ÷ (6x³ − x). Which line is the first to go wrong?
Dev’s horizontal asymptote of f(x) = (3x² + 5)/(6x³ − x)
- 1top: 3x², bottom: 6x³
- 2leading coefficients 3 and 6
- 3degrees equal, so the ratio decides: 3 ÷ 6 = 0.5
- Answerhorizontal asymptote y = 0.5
4 of 6
What is the horizontal asymptote value of f(x) = (3x² + 5) ÷ (6x³ − x)?
5 of 6
Noor tested the two sides of the vertical asymptote of f(x) = (x + 5) ÷ (x − 2). Which line is the first to go wrong?
Noor’s two sides of the asymptote of f(x) = (x + 5)/(x − 2)
- 1x = 2: bottom 0, top 7 ≠ 0
- 2x = 2.1: 7.1 ÷ 0.1 = 71
- 3x = 1.9: 6.9 ÷ 0.1 = 69
- Answerboth sides run to +∞
6 of 6
What is f(1.9) for f(x) = (x + 5) ÷ (x − 2)? Type it as a decimal.
Pick the Move
Decide which tool the question calls for. Neither one has to be carried all the way through.
1 of 2
You are asked for every vertical asymptote of f(x) = (x² − x − 6) ÷ (x² − 4). Which first move?
2 of 2
One question asks where the graph of f(x) = (5x + 1) ÷ (2x − 7) breaks; another asks what it does far out to the right. Which tool answers each?
Same Asymptote?
Judge these without drawing anything. The highest power on each half decides a far end; the top at a zero of the bottom decides a break.
1 of 2
Select every function whose graph has the horizontal asymptote y = 2.
2 of 2
Select every function whose graph has a vertical asymptote at x = 3.
Step 6: Quick Check
Show what you know.
Question 1 of 2
Limit of (8x + 3) ÷ (4x − 1) as x grows without bound?
Question 2 of 2
A denominator is zero and the numerator is not. What is at that point?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which number k does f(x) = (kx + 5) ÷ (2x − 3) have the horizontal asymptote y = 4?
Stretch 2 of 2
f(x) = (3x + 6) ÷ (x − 2) has a vertical asymptote at x = 2. Changing only the 6, what number in its place turns that break into a hole?
What You Learned
- A limit at infinity gives the horizontal asymptote.
- Compare degrees: equal gives a ratio, smaller top gives 0, larger top gives none.
- A vertical asymptote needs a zero denominator and a non-zero numerator.
- Factor before declaring an asymptote: a canceling factor makes a hole.
For the grown-up
The misconception this lesson targets is that every zero of the denominator is a vertical asymptote. It is an easy belief to hold, because it is right most of the time and because the fraction really is undefined at every one of those x-values — but undefined is not the same as unbounded. Where the numerator vanishes at the same x, the two share a factor, it cancels, and the graph has a single missing point at a perfectly ordinary height. A student who does not factor first will report a wall where there is a pinprick, and the sketch that follows will be wrong everywhere near it. The cure is a habit rather than a rule: factor both halves before saying anything, and work the numerator out at each zero of the denominator. Its cousin is the belief that a horizontal asymptote is a barrier the curve may never touch or cross. It describes the far ends only; plenty of curves cross their own horizontal asymptote near the origin and then settle onto it from the other side.
Two smaller slips ride along with this one. The first is taking the ratio of the leading coefficients whatever the degrees are: 3x² over 6x³ is not one half of anything, because a cube outgrows a square no matter what numbers stand in front of them, and the honest move — dividing top and bottom by the highest power of x — reports all three cases without any rule to remember. The second is testing only one side of a vertical asymptote. Just to the left of x = 2 the quantity x − 2 is a small negative, and a student who writes 0.1 where −0.1 belongs will have the curve climbing both sides of a wall it actually falls down on one of them. Asking out loud "is the bottom positive or negative just there?" catches nearly all of it, and it is the same question the sign of a limit will keep asking for the rest of the course.