Applications: Limits

The formula is a quotient, and at r = 1 the top and the bottom are both zero. That is an indeterminate form: it does not say the value is 1, or 0, or that there is none. The way through is to factorize the top, cancel the factor that is zero, and read the limit off what is left, which turns out to be the sum of the twenty payments.

1. Substitute r = 1 first, to see what happens. The top is 1²⁰ − 1 = 0 and the bottom is 1 − 1 = 0, so the formula asks for 00. That is an indeterminate form, and it settles nothing; the spreadsheet reports an error because it is being asked to divide by zero.
2. Work the value out either side of 1 instead. At r = 0.999 the formula gives $3962.23 and at r = 1.001 it gives $4038.23. Closer in, r = 0.9999 gives $3996.20 and r = 1.0001 gives $4003.80. The values are closing in on $4000 from both sides.
3. Now do the algebra that explains it. The top factorizes: r²⁰ − 1 = (r − 1)(r¹⁹ + r¹⁸ + … + r + 1). On the way to 1 the growth factor is near 1 but never equal to it, so r − 1 is not zero and may be canceled: T(r) = 200(1 + r + r² + … + r¹⁹).
4. The canceled form is a sum of twenty terms, and the limit laws let a limit pass through a sum. Each term tends to 1, so the sum tends to 20, and lim_{r → 1} T(r) = 200 × 20 = 4000.
5. (a) The limit is $4000. That is the twenty payments of $200 with no growth added to any of them, which is what a year of no growth should give. The formula has no value at r = 1, but the value it closes on is the sensible one, so the spreadsheet should use 200 × 20 there.
6. (b) With no growth, n payments are worth 200n dollars. Solve 200n = 6000, so n = 30. Check: 30 × 200 = 6000, so the club would need ten years more than the twenty it has planned.

answer(a) lim_{r → 1} T(r) = $4000, the twenty payments of $200 with no growth at all, so the spreadsheet should use 200 × 20 for a year of no growth; (b) 30 payments

techniqueIndeterminate Forms · The Idea of a Limit · The Limit Laws

A charge that steps up is a function given by a different rule on each piece of its domain. At a step the two pieces have to be looked at separately: the limit from the left uses the rule for shorter stays and the limit from the right uses the rule for longer ones. A two-sided limit exists only when those two numbers agree.

1. Fix which rule applies on each side of 3 hours. A stay of 2.9 hours, of 2.99 hours or of 2.999 hours is a stay of up to 3 hours, so the charge is $4 every time.
2. A limit is about values near the point, not at it. The charge is the constant 4 all the way up to 3, and the limit of a constant is that constant, so lim_{x → 3⁻} C(x) = 4.
3. Just above 3 hours the second rule applies and the charge is the constant $7: at 3.001 hours it is $7, and at 3.1 hours it is $7. So lim_{x → 3⁺} C(x) = 7.
4. (a) The left limit is 4 and the right limit is 7. A two-sided limit exists only when the one-sided limits agree, so lim_{x → 3} C(x) does not exist. The charge at exactly 3 hours is $4, because a stay of 3 hours counts as a stay of up to 3 hours.
5. At the six-hour mark the same reading gives lim_{x → 6⁻} C(x) = 7 and lim_{x → 6⁺} C(x) = 12, so the graph jumps again, this time by 12 − 7 = 5.
6. (b) The first driver pays $7 and the second pays $12, so the second pays $5 more. Two minutes of parking separate them, and the price of those two minutes is the size of the jump.

answer(a) lim_{x → 3⁻} C(x) = 4 and lim_{x → 3⁺} C(x) = 7; they disagree, so lim_{x → 3} C(x) does not exist, while C(3) = 4; (b) $5 more

techniqueOne Sided Limits · The Idea of a Limit

The price is given by one rule for light parcels and another for heavy ones, and the two rules meet at 10 kg. No jump means the price approached from below and the price approached from above are the same number, so the two one-sided limits give an equation for k.

1. The join is at w = 10. Just below it the first rule applies, so lim_{w → 10⁻} C(w) = 5 + 2 × 10 = 25 dollars. The charge at exactly 10 kg is the same $25, because w ≤ 10 uses the first rule.
2. Just above the join the second rule applies, so lim_{w → 10⁺} C(w) = k + 1.5 × 10 = k + 15 dollars.
3. No jump means those two prices are equal: k + 15 = 25.
4. (a) k = 25 − 15 = 10. Check: the second rule at 10 kg gives 10 + 15 = $25, the same as the first rule, so the two pieces meet.
5. (b) A parcel of 16 kg is over 10 kg, so the second rule applies: C(16) = 10 + 1.5 × 16 = 10 + 24 = $34. Check by adding on from the join: above 10 kg each extra kilogram adds $1.50, and 25 + 6 × 1.50 = $34.

answer(a) k = 10; (b) $34

techniqueOne Sided Limits · The Limit Laws

The share is one linear expression divided by another, so the long run is settled by the terms in x rather than by the numbers the towns start with. Dividing the top and the bottom by x makes that plain: the constant terms turn into fractions that fade to zero, and the limit laws then give the limit as a quotient of the two growth rates.

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.
2. For the long run, divide every term by x: f(x) = 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: lim_{x → ∞} f(x) = 0 + 8000 + 2000 = 25.
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%.
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.
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%.

answer(a) 25, that is 40%; (b) year 31

techniqueLimits at Infinity · The Limit Laws

The model is a constant minus a decaying exponential. The whole of the long-run behavior sits in that exponential term, which tends to zero, so the limit laws hand back the constant. The second part reverses the model: a stated speed gives an equation in the exponential, and logarithms solve it.

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 lim_{x → ∞} e^−x/6 = 0.
2. The limit laws pass through the bracket and the multiple: lim_{x → ∞} 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.
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.
5. Take logarithms of both sides: −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.

answer(a) lim_{x → ∞} v(x) = 60 m/s, the terminal speed the fall settles at and never quite reaches; (b) 13.8 seconds

techniqueLimits at Infinity · The Limit Laws

The sine in the model has no limit at the tip: it swings between −1 and 1 faster and faster. What it does have is a fixed pair of bounds, and multiplying those bounds by x² traps the error between two curves that both tend to zero. That is the squeeze theorem, and it also answers the second part, since the upper bound is a promise about the whole of the region near the tip.

1. The sine of any number lies between −1 and 1, so −1 ≤ sin1x ≤ 1 for every x except 0. The awkward part of the model is inside those bounds, however fast it swings.
2. Multiply the inequality through by x². That is a positive number for x ≠ 0, so the inequality signs stay as they are: −x² ≤ e(x) ≤ x².
3. Both bounding curves have the same limit at the tip: lim_{x → 0} (−x²) = 0 and lim_{x → 0} x² = 0. The error is trapped between two curves that arrive at the same place.
4. (a) By the squeeze theorem, lim_{x → 0} e(x) = 0. The ripple dies away at the tip: the model has no value at x = 0, but every value near it is near 0, so the surface is true there.
5. (b) The same bound answers this. For 0 < x ≤ 0.3, the error is at most x², and x² is at most 0.3² = 0.09. So the surface is out by at most 0.09 mm. Check: at x = 0.3 the model gives e(0.3) = 0.09 sin 3.33 = −0.017 mm, which is well inside the bound.

answer(a) lim_{x → 0} e(x) = 0 mm, so the ripple dies away at the tip; (b) at most 0.09 mm

techniqueThe Squeeze Theorem · The Idea of a Limit

Both distances carry the same radius, so the ratio of the two depends on the angle alone. Written in terms of half the angle it is exactly sin uu, whose limit at zero is the standard one. The limit says the two measurements agree for a small enough angle; putting a particular angle in says by how much they still differ.

1. Write the ratio and cancel the radius: 2 × 7000 sinθ27000θ = 2sinθ2θ. The size of the orbit has gone, and only the angle is left.
2. Put u = θ2, so that θ = 2u and the ratio is 2sin u2u = sin uu. As θ → 0 the half-angle u → 0 as well.
3. The standard limit for an angle in radians is lim_{u → 0} sin uu = 1, so the ratio of the chord to the arc tends to 1.
4. (a) The limit is 1. For a small enough angle the straight line and the arc agree as closely as you please, which is why a short hop across an orbit may be measured either way.
5. (b) The arc is 7000 × 0.1 = 700 km. The chord is 14000 sin 0.05 = 14000 × 0.0499792 = 699.71 km.
6. The difference is 700 − 699.71 = 0.29 km, about 290 m, which is 0.04% of the arc. Check with the ratio itself: sin 0.050.05 = 0.99958, and 700 × 0.99958 = 699.71 km.

answer(a) 1; (b) the arc is 700 km, the chord is 699.71 km, and the difference is 0.29 km, about 290 m

techniqueThe Limit of sin x over x · The Idea of a Limit

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With the speed fixed, the formula is a constant divided by the grip. The grip cannot be negative, so the only approach to zero is from above, and along it the top stays put while the bottom shrinks: the distance passes every length. The target in part (b) is then an inequality in μ.

1. Put the numbers in: d(μ) = 20²2 × 10 × μ = 40020μ = 20μ meters. The speed and g are fixed, so the whole of the distance is settled by the grip.
2. On dry tarmac, d(0.8) = 200.8 = 25 m. On ice, d(0.1) = 200.1 = 200 m, eight times as far.
3. Now let the grip fall toward zero from above. The top stays at 20 while the bottom tends to 0 through positive values, so the quotient grows without bound: d(0.01) = 2000 m, and d(0.001) = 20000 m.
4. (a) The distances are 25 m and 200 m, and lim_{μ → 0⁺} d(μ) does not exist: the stopping distance passes every length you name. On a road with no grip at all the car would not stop.
5. (b) The target is 20μ ≤ 50. The grip is positive, so multiplying both sides by μ keeps the inequality the way round it is: 20 ≤ 50μ, and μ ≥ 0.4.
6. The least grip is 0.4. Check: d(0.4) = 200.4 = 50 m exactly, while a slightly smaller grip of 0.39 gives 51.3 m, which misses the target.

answer(a) 25 m on dry tarmac and 200 m on ice; lim_{μ → 0⁺} d(μ) does not exist, because the distance passes every length; (b) a grip of at least 0.4

techniqueOne Sided Limits · The Idea of a Limit

The ratio of the depth to the square of the distance is 00 at the axis, and the surd is what blocks the canceling. Multiplying by the conjugate turns the top into a difference of two squares, the surd disappears from it, and the a² that made the bottom zero cancels. What is left may be substituted into directly.

1. Substituting a = 0 gives 2 − √40 = 00, an indeterminate form. The surd is what blocks any canceling, so clear it first.
2. Multiply the top and the bottom by the conjugate 2 + √4 − a². The top becomes a difference of two squares: (2 − √4 − a²)(2 + √4 − a²) = 4 − (4 − a²) = a².
3. So for a ≠ 0, s(a)a² = a²a²(2 + √4 − a²) = 12 + √4 − a². The a² that made the bottom zero has canceled, and the surd that is left is harmless at a = 0.
4. Substituting is safe now: 12 + √4 − 0 = 14.
5. (a) The limit is 14, that is 0.25, per meter. It says that near the axis the depth is about a²4, which is the makers' rule, and it shows where the rule comes from: 12 × 2, one over twice the radius.
6. (b) The rule gives 0.6²4 = 0.364 = 0.09 m. The exact depth is 2 − √4 − 0.36 = 2 − √3.64 = 0.0921 m. The rule is about 2 mm shallow, which for a cooker's dish is close enough to build to.

answer(a) lim_{a → 0} s(a)a² = 14, that is 0.25 per meter, which is where the makers' rule comes from; (b) the rule gives 0.09 m against an exact 0.0921 m, so it is about 2 mm shallow

techniqueIndeterminate Forms · The Limit Laws

Near zero the trigonometric functions flatten into polynomials: tanθ and sinθ both behave like θ, and cosθ like 1 − θ²2. Both angles here are a few hundredths of a radian or less, so the polynomials do the arithmetic. The second part is the case that needs the quadratic term, because the constant term cancels.

1. The height is 800tanθ with θ = 0.02 radians. The angle is small, and for a small angle in radians tanθ ≈ θ.
2. So the height is about 800 × 0.02 = 16 m.
3. (a) The mast is 16 m tall. The exact value is 800tan 0.02 = 16.0021 m, so the approximation is about 2 mm short over 16 m, which is well inside what a tape would settle.
4. For part (b), θ = 86400 = 0.00125 radians, smaller still. With cosθ ≈ 1 − θ²2, the bracket becomes 1 − cosθ ≈ θ²2 = 0.00125²2 = 7.8125 × 10⁻⁷.
5. The drop is 6400 × 7.8125 × 10⁻⁷ = 0.005 km.
6. (b) The drop is 5 m over 8 km. Check with the exact cosine: 6400(1 − cos 0.00125) = 0.0050000 km, the same to the nearest millimeter, because the next term of the approximation is of size θ⁴24, about 10⁻¹³.

answer(a) 16 m, about 2 mm short of the exact 16.0021 m; (b) a drop of 5 m over 8 km

techniqueSmall-Angle Approximations · The Limit of sin x over x

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