What happens in the long run? The cost of each item when a factory makes more and more of them, the amount of a medicine left after many hours, the speed a falling raindrop settles at. All of these are limits as x → ∞: x grows past every number, and we ask where f(x) is heading. One idea is behind every exercise in this chapter: far out, only the fastest-growing part of an expression matters. You will learn to spot that part, divide by it and read off the answer, and to handle the few places where that idea needs care: minus infinity, square roots, and ∞ − ∞.
By the end you can
- Say what lim f(x) = L as x → ∞ means, and check a guess with a table of values.
- Find the limit of any ratio of polynomials at +∞ or −∞, by dividing and by comparing degrees.
- Get the sign right with square roots at −∞, and undo ∞ − ∞ with the conjugate.
- Rank logarithms, powers and exponentials by how fast they grow, and use the ranking to find limits.
x → ∞ (read "x tends to infinity") means that x takes larger and larger values, 10, 100, 1000 and on, passing every number you could name. We never plug in ∞: infinity is not a number, and 1/∞ is not a calculation. We watch what f(x) does while x keeps growing. Take a factory that pays 5000 to set up a machine and then 2 for each item it makes. Making x items costs 5000 + 2x, so the cost per item is C(x) = (5000 + 2x)/x.
| items x | 10 | 100 | 1000 | 10 000 | 1 000 000 |
|---|---|---|---|---|---|
| cost per item C(x) | 502 | 52 | 7 | 2.5 | 2.005 |
The cost per item closes in on 2: the setup cost is shared by so many items that each one carries almost none of it. Algebra says the same: C(x) = 5000/x + 2, and 5000/x shrinks toward 0. We write lim C(x) = 2 as x → ∞. The line y = 2 is called a horizontal asymptote: the graph gets as close to it as you like, and here never touches it.
x → −∞ is the same idea to the left: x = −10, −100, −1000 and on. And when f(x) itself grows past every number, we write lim f(x) = ∞. That is not a value: it says the values run away, and on that side the graph has no horizontal asymptote.
Every limit in this chapter is assembled from three facts. Each one can be checked with a single number.
| fact | why | check at x = 1000 |
|---|---|---|
| c / xᵏ → 0, for any number c and any power k > 0 | a fixed number divided by a huge one is tiny | 5 / x² = 0.000005 |
| xᵏ → +∞ as x → +∞, for k > 0 | a power of a huge number is even larger | x² = 1 000 000 |
| a constant c stays c | it does not depend on x | 7 = 7 |
At −∞ the second fact depends on the power. An even power is positive, so it still goes to +∞: (−1000)² = 1 000 000. An odd power keeps the minus sign and goes to −∞: (−1000)³ = −1 000 000 000. The first fact holds at −∞ too: 5/(−1000) = −0.005 is still tiny.
Arithmetic with infinity
Combining the blocks, some results are clear and some are not. When both parts push the same way the answer is decided. When they pull against each other it is not: the form alone tells you nothing, and you must rewrite the expression first. These undecided forms are called indeterminate forms.
| form | result | why |
|---|---|---|
| ∞ + ∞ | ∞ | both grow |
| c · ∞ | ±∞ | +∞ if c > 0, −∞ if c < 0 (c ≠ 0) |
| c / ∞ | 0 | the first building block |
| ∞ · ∞ | ∞ | both grow |
| ∞ − ∞ | undecided | x² − x → ∞, but (x + 1) − x = 1 and x − x² → −∞ |
| ∞ / ∞ | undecided | 2x/x → 2, x²/x → ∞, x/x² → 0 |
| 0 · ∞ | undecided | (1/x) · x = 1, (1/x) · x² → ∞ |
Almost every exercise starts as ∞/∞ or ∞ − ∞. The rest of the chapter is the toolbox for turning those into something decided.
Take p(x) = 2x³ − 100x² + 5. As x → ∞ the first term goes to +∞ and the second to −∞: ∞ − ∞, undecided. Look at the sizes instead. At x = 1000 the three terms are 2 000 000 000, −100 000 000 and 5. The first is twenty times the second, and the gap keeps widening, because x³ is x times x², and x keeps growing. The term with the highest power is called the leading term, and far out it is the only one that matters.
At −∞ the bracket still tends to 2, but x³ → −∞, so p(x) → −∞. The rule: a polynomial at ±∞ goes where its leading term goes. Only two things decide it: the sign of the leading coefficient, and, at −∞, whether the degree is even or odd. So −x⁴ + 50x³ → −∞ on both sides, and 3x⁵ − x → +∞ at +∞ but −∞ at −∞.
A fraction of two polynomials is ∞/∞ (or something over ∞), undecided. The method has four steps, and it always works:
| step | what you do | (3x² − 20x)/(x² + 5) |
|---|---|---|
| 1 | Find the highest power of x in the denominator. | x² |
| 2 | Divide every term, top and bottom, by that power. The fraction's value does not change: dividing top and bottom by the same thing is multiplying by 1. | (3 − 20/x) / (1 + 5/x²) |
| 3 | Every term of the form c/xᵏ goes to 0. | (3 − 0) / (1 + 0) |
| 4 | Read off what is left: a number, or something that still grows. | 3 |
Which of three things happens depends only on how the degree of the top (its highest power, m) compares with the degree of the bottom (n).
Case 1: the same degree gives the ratio of the leading coefficients
The −20x is not small at first: at x = 5 the fraction is (75 − 100)/30 ≈ −0.83, nowhere near 3. But it grows only like x while 3x² grows like x², so far out it no longer counts. The figure below shows exactly this.
Case 2: a bigger degree below gives 0
The bottom wins the race: it grows like x², the top only like x, so the fraction behaves like 3x/x² = 3/x, which is crushed to 0.
Case 3: a bigger degree on top gives ±∞
| degrees | limit at ±∞ | example |
|---|---|---|
| m < n | 0 | (3x + 2)/(x² + 1) → 0 |
| m = n | a / b | (3x² − 20x)/(x² + 5) → 3 |
| m > n | ±∞: the sign of (a/b) · x^(m − n) on that side | (x² − 8x)/(2x + 1) → +∞ at +∞, −∞ at −∞ |
Press play, or ⏭, to zoom out: the window grows from |x| ≤ 10 to |x| ≤ 10 000. The blue curve is the function; the dashed amber curve keeps only its dominant term, the fastest-growing part of the top over the fastest-growing part of the bottom. Near 0 they disagree, sometimes wildly, but as the window grows the blue curve melts into the amber one and f / dominant closes in on 1. So the limit of f is the limit of its dominant term, which is easy: a number (the green line, a horizontal asymptote), or a curve that runs off to ±∞. Switch to x → −∞ and watch the square root and the conjugate change their answer.
Roots change one thing. To take x² out of a square root you use √(x²) = |x|, the absolute value (the number without its sign), because a square root is never negative. For x → +∞, x is positive and |x| = x, so nothing changes. For x → −∞, x is negative, so |x| = −x. Forgetting that minus sign is the most common mistake in this whole chapter.
At +∞ the same steps, with |x| = x, give +2. One function, two horizontal asymptotes: y = 2 on the right and y = −2 on the left. A sign check with one number catches the mistake every time: at x = −1000 the top is about 2000 (a root is positive) and the bottom is −997, so the fraction is about −2.
√(x² + 6x) − x as x → +∞ is ∞ − ∞. Here the leading-term idea alone fails: inside the root the leading term is x², √(x²) = x, and x − x = 0, but the answer is not 0. The two big terms almost cancel, and what decides is the small leftover. Multiplying by the conjugate (the same two terms with the sign between them flipped) turns the difference into a fraction, and fractions we know how to handle.
Check with a number: at x = 1000, √(1 006 000) ≈ 1002.996, and minus 1000 leaves 2.996. A rule of thumb: for large x, √(x² + bx) ≈ x + b/2, because (x + b/2)² = x² + bx + b²/4 is almost the same thing under the root. At −∞ there is nothing to undo: √(x² + 6x) is positive and −x is positive too, so the sum is ∞ + ∞ = +∞. Undecided forms must be rewritten, but first make sure the form really is undecided.
Beyond polynomials the same idea, divide by the fastest-growing term, needs a ranking of how fast things grow as x → ∞.
| x | ln x | x² | 2ˣ |
|---|---|---|---|
| 10 | 2.3 | 100 | 1024 |
| 100 | 4.6 | 10 000 | 1.27 · 10³⁰ |
| 1000 | 6.9 | 1 000 000 | 1.07 · 10³⁰¹ |
Why the exponential wins: going from x to x + 1 multiplies 2ˣ by 2, every single time. The same step multiplies x² by ((x + 1)/x)², which is 1.21 at x = 10 and only 1.0201 at x = 100, closer and closer to 1. A fixed doubling at every step beats a growth factor that fades toward 1. The logarithm is the slow one because it undoes the exponential: for ln x to reach 100 you need x = e¹⁰⁰, a number with 44 digits.
At −∞ an exponential with base above 1 goes to 0: 2⁻¹⁰⁰⁰ = 1/2¹⁰⁰⁰. So the same function at −∞ becomes (0 + x)/(0 + 1), which follows x down to −∞. Switch the figure to x → −∞ to see it. In the same way e⁻ˣ → 0 as x → ∞: that is the shape of a cooling cup, of radioactive decay and of a charging capacitor, all of which settle on a final value.
(1 + 1/x)ˣ looks like 1^∞: the base heads for 1, and 1 to any power is 1, but the exponent grows without bound, and a number even slightly above 1, raised to a huge power, is huge. The two pull against each other, and the result is the number e ≈ 2.71828, which you met in the exponents chapter as the limit of compound interest. A substitution turns it into a whole family.
So (1 + 3/x)ˣ → e³ ≈ 20.09 and (1 − 1/x)ˣ → e⁻¹ ≈ 0.368 (for a negative k the same steps work with u = x/k → −∞, because (1 + 1/u)ᵘ tends to e from that side too). In money terms: 100 % yearly interest paid in x tiny instalments grows 1 into e, and 300 % into e³.
When the top degree is exactly one more than the bottom, the limit is ±∞, but the graph still settles along a line, a slanted (oblique) asymptote. Polynomial division finds it. Dividing x² − 8x by 2x + 1 gives the quotient x/2 − 17/4 and the remainder 17/4, so (x² − 8x)/(2x + 1) = x/2 − 17/4 + (17/4)/(2x + 1). The last part is a number over something that grows, so it goes to 0, and far out the graph runs along the line y = x/2 − 17/4.
| you see | what to do |
|---|---|
| nothing undecided (∞ + ∞, c/∞, c · ∞ …) | read it off directly |
| a polynomial | follow its leading term |
| a ratio of polynomials (∞/∞) | divide every term by the highest power in the denominator, or compare the degrees |
| a square root of x² + … | take out |x|; at −∞ write |x| = −x |
| a root minus a term (∞ − ∞) | multiply by the conjugate, then divide |
| exponentials, powers and logarithms mixed | divide by the fastest-growing term, using the ranking |
| (1 + k/x)ˣ, the form 1^∞ | it tends to eᵏ |
1. lim (4x³ − x + 7) as x → −∞. A polynomial: it follows 4x³. An odd power of a large negative number is large and negative, times 4: −∞.
2. lim (6x² − x)/(3 − 2x²) as x → ∞. Same degree. The leading term below is −2x², even though it is written last: 6/(−2) = −3.
3. lim (x + 5)/(x³ − 1) as x → −∞. The bottom has the bigger degree: 0.
4. lim 2x³/(x² + 1) as x → −∞. The top has the bigger degree. Leading terms: 2x³/x² = 2x, which goes to −∞ when x does.
5. lim √(9x² + x)/(3x − 4) as x → −∞. Take out |x| = −x: −√(9 + 1/x)/(3 − 4/x) → −3/3 = −1.
6. lim (√(x² + 4x) − x) as x → ∞. ∞ − ∞: the conjugate gives 4x/(√(x² + 4x) + x); divide by x: 4/(√(1 + 4/x) + 1) → 4/2 = 2.
7. lim (x³ + 5ˣ)/(2 · 5ˣ − x) as x → ∞. The fastest term is 5ˣ; divide by it: (x³/5ˣ + 1)/(2 − x/5ˣ) → (0 + 1)/(2 − 0) = 1/2.
8. lim (1 + 2/x)ˣ as x → ∞. The form 1^∞ with k = 2: e² ≈ 7.389.
9. lim (ln x)/√x as x → ∞. A logarithm over a power: by the ranking, 0. (At x = 10⁶ it is already 13.8/1000 = 0.014.)
A round of 10 questions
Answer as many as you can. Mistakes come back at the end for a review round.
Every exercise is made on the spot, so a new round is never the same. The computer picks the answer first and builds the question around it, working with exact fractions, so the answer key is always right. Answers are read exactly: 0.75, 3/4 and 6/8 count as the same number (where lowest terms are asked for, 6/8 just gets a reminder). A wrong answer shows the worked solution line by line, in the order the lesson does it; the new number in each line is coloured.
| Wrong | Right | Why |
|---|---|---|
| ∞ − ∞ = 0 | rewrite: factor, or use the conjugate | √(x² + 6x) − x → 3, not 0 |
| ∞ / ∞ = 1 | divide by the fastest-growing term | 2x/x → 2, x²/x → ∞ |
| √(x²) = x when x → −∞ | √(x²) = |x| = −x there | a square root is never negative |
| dividing only the top by xⁿ | divide top and bottom by the same power | otherwise the fraction changes value |
| taking the first term written as the leading one | look for the highest power | in 3 − 2x² the leading term is −2x² |
| 1^∞ = 1 | (1 + k/x)ˣ → eᵏ | the base and the exponent race each other |
| 2ˣ → ∞ as x → −∞ | 2ˣ → 0 as x → −∞ | 2⁻¹⁰ = 1/1024 |
Key ideas
- x → ∞ means x grows past every number. Infinity is never plugged in; you watch where f(x) heads.
- Far out, only the fastest-growing term of each part matters.
- Ratio of polynomials: same degree gives the ratio of the leading coefficients, a bigger bottom gives 0, a bigger top gives ±∞.
- To show it, divide every term by the highest power in the denominator; every c/xᵏ goes to 0.
- √(x²) = |x|: at −∞ that is −x, and the sign of the answer flips.
- ∞ − ∞, ∞/∞, 0 · ∞ and 1^∞ are undecided: rewrite before concluding (factor, conjugate, divide).
- ln x ≪ xᵖ ≪ aˣ: logarithms lose to powers, powers lose to exponentials; and aˣ → 0 at −∞.
- (1 + k/x)ˣ → eᵏ.