Far out to the right
A limit as asks what value f(x) settles on as x grows without bound. Take . At x = 0 it is 1, at x = 1 it is 1.5, at x = 10 it is , at x = 100 it is , and at x = 1000 it is .
The values climb toward 2 and flatten off. They never reach it: , which is negative for every x > −1, so the curve stays below 2. The gap shrinks to 0 as x grows, so the curve gets as close to 2 as you like.
The graph of for x from 0 to 20, with the line y = 2 dashed. The curve starts at 1, climbs quickly, then flattens just below the line: at x = 20 it is , about 1.95.
Divide by the highest power of x
The highest power of x in is x itself. Dividing the numerator and the denominator by the same nonzero number leaves a fraction unchanged, so for , .
Now every term that is not a plain number has a power of x underneath it. As x grows, shrinks to 0: at x = 1000 it is 0.001. So the numerator tends to 2 and the denominator tends to 1.
The denominator tends to 1, which is not 0, so the limit of the quotient is the quotient of the limits, and the limit of as is . The line y = 2 is a horizontal asymptote of the curve: the line it settles onto far out.
The ratio of the leading coefficients
After the division, only the terms that held the highest power survive, and they are the leading coefficients: the 2 of 2x and the 1 of x. So when the numerator and the denominator have the same degree, the limit is the ratio of their leading coefficients.
For , divide by x: . At x = 1000 the fraction is .
For , the highest power is , so divide by : . At x = 100 it is . The line y = 3 is its horizontal asymptote.
When the degrees differ
For , the highest power is . Divide by it: . The numerator tends to 0 and the denominator to 1, so the limit is 0, and the x-axis is the horizontal asymptote. At x = 100 the fraction is .
For , divide by : . The numerator tends to 1 but the denominator tends to 0 through positive values, and 1 divided by a smaller and smaller positive number grows without bound. At x = 100 the fraction is , and at x = 1000 it is 997.0. There is no finite limit and no horizontal asymptote.
So compare the degrees. Equal degrees give the ratio of the leading coefficients; a higher degree underneath gives 0; a higher degree on top gives no finite limit.
at x = 16 the terms 5x − 3 still make 13.2% of the numerator and f = 2.341: zoom out and the lower powers lose their share, so f → 2x²/x² = 2
Zoom out until the lower terms are under 0.5% of the numerator at the edge
The curve is , and the dashed line is y = 2. At a window of about the branches beside x = −2 and x = 2 fill the picture. Zoom out: at x = 1000 the terms 5x − 3 are about a quarter of one percent of the numerator, and the curve lies along y = 2.
Far out to the left
As , shrinks to 0 from below, so the same division gives the same limit: . The gap is now positive, because x + 1 is negative, so the curve approaches 2 from above: at x = −1000 the value is .
The usual mistakes
Answering 0 because the terms shrink to 0. Only those pieces vanish; the leading coefficients are left, and tends to 2, not 0.
Adding the leading coefficients. tends to , not 8 + 1 = 9: the fraction divides them.
Reading the limit off the constant terms. For , the 1 and the 1 are what stops mattering far out; is the value at x = 0, not the limit.
Dividing the top and the bottom by different powers of x. Both must be divided by the same power, or the fraction changes.
A district in the long run, and a falling skydiver
In the first application below, one town’s share of a district is a quotient of two linear expressions in the number of years, and its long-run value comes from dividing by x. In the second, the speed involves , which tends to 0 as x grows, just as does: at x = 24 it is , about 0.018.
Worked example: Two Towns in One District: The Share of the People One Town Keeps in the Long Run
Question A district has two towns. Alder has 12000 people and grows by 800 a year; Brook has 8000 people and grows by 1200 a year. After x years Alder's share of the district is f(x) = 12000 + 800x20000 + 2000x. (a) Find limx → ∞ f(x), as a percentage. (b) Find the first whole year in which Alder's share is below 45%.
1.The district total is (12000 + 800x) + (8000 + 1200x) = 20000 + 2000x people, which is the bottom of the share, and Alder's people are the top.
The district holds 20000 + 2000x people, and Alder holds 12000 + 800x of them, which is 60% today. 2.For the long run, divide every term by x: f(x) = 12000x + 80020000x + 2000.
Divide every term by x: the share is 12000x + 80020000x + 2000. 3.As x → ∞ both 12000x and 20000x tend to 0. The bottom tends to 2000, which is not zero, so the limit laws let the limit pass through the quotient: limx → ∞ f(x) = 0 + 8000 + 2000 = 25.
As x → ∞ the two fractions tend to 0, and the bottom tends to 2000, so the share tends to 8002000. 4.(a) 25 is 40%. In the long run Alder holds 40% of the district, because what settles the share is the two growth rates, 800 and 1200 a year, and not the two starting numbers. Check: after a thousand years the share is 8120002020000, which is 40.2%.
(a) limx → ∞ f(x) = 25, that is 40%. The two growth rates settle the long-run share. 5.For part (b), solve 12000 + 800x20000 + 2000x < 0.45. The bottom is positive, so multiplying both sides by it keeps the inequality: 12000 + 800x < 9000 + 900x, which gives 3000 < 100x and x > 30.
For the target, 12000 + 800x < 9000 + 900x, which gives 3000 < 100x and x > 30. 6.(b) The first whole year is year 31. Check: at x = 30 the share is 3600080000 = 45% exactly, and at x = 31 it is 3680082000, which is 44.9%.
(b) Year 31 is the first whole year below 45%: the share is 44.9%, and at year 30 it is exactly 45%.
Answer: (a) 25, that is 40%; (b) year 31
Common mistakes
- Reading the limit off the constant terms as 1200020000 = 60%, the share the district starts with. Those constants keep their size while the towns grow, so their share of the total fades; the limit is settled by the coefficients of x.
- Taking x > 30 to mean year 30. The share at x = 30 is exactly 45%, which is not below 45%, so the first year that meets the condition is year 31.
Worked example: A Skydiver Before the Parachute Opens: The Speed the Fall Settles At
Question A skydiver's speed x seconds after the jump is modeled by v(x) = 60(1 − e−x/6) meters per second. (a) Find limx → ∞ v(x) and say what it means. (b) Find the time at which the speed reaches 54 m/s, which is 90% of that limit. Give it to three significant figures.
1.As x grows, −x6 is a larger and larger negative number and e raised to it is smaller and smaller: at x = 24 it is e−4 ≈ 0.018, and at x = 120 it is e−20, about 2 × 10−9. So limx → ∞ e−x/6 = 0.
As x grows, e−x/6 falls toward 0: at 24 seconds it is 0.018 and the speed is 58.9 m/s. 2.The limit laws pass through the bracket and the multiple: limx → ∞ 60(1 − e−x/6) = 60(1 − 0) = 60.
The limit laws pass through the bracket: limx → ∞ 60(1 − e−x/6) = 60(1 − 0) = 60. 3.(a) The limit is 60 m/s. The fall settles at 60 m/s, the terminal speed, where the air resistance balances the weight. It is never quite reached, because e−x/6 is greater than 0 for every x: after 24 seconds the speed is 58.9 m/s.
(a) The fall settles at 60 m/s, the terminal speed. It is approached and never quite reached. 4.For part (b), put v(x) = 54: 60(1 − e−x/6) = 54, so 1 − e−x/6 = 0.9 and e−x/6 = 0.1.
For part (b), 60(1 − e−x/6) = 54 gives 1 − e−x/6 = 0.9, so e−x/6 = 0.1. 5.Take logarithms of both sides: −x6 = ln 0.1 = −ln 10, so x = 6 ln 10.
Take logarithms: −x6 = ln 0.1 = −ln 10, so x = 6 ln 10. 6.(b) x = 6 ln 10 = 13.8 seconds to three significant figures. Check: v(13.8) = 60(1 − e−2.3) = 53.98 m/s, which is the 54 m/s asked for.
(b) x = 6 ln 10 = 13.8 seconds. The check agrees: v(13.8) = 53.98 m/s.
Answer: (a) limx → ∞ v(x) = 60 m/s, the terminal speed the fall settles at and never quite reaches; (b) 13.8 seconds
Common mistakes
- Solving 60(1 − e−x/6) = 60 for the moment the terminal speed is reached. That needs e−x/6 = 0, and no value of x makes an exponential zero. The terminal speed is a limit, approached and not reached, which is why the question asks for 90% of it.
- Writing e−x/6 = 0.1 and then x = 6 ln 0.1 = −13.8. The minus sign belongs to the exponent: −x6 = ln 0.1, so x = −6 ln 0.1 = 6 ln 10. A negative time would be before the jump, which the situation rejects.