Where a curve can break
A function is continuous at a when f(a) exists, the limit as exists, and the two are equal. A point where that fails is a discontinuity. There are three types, told apart by the two one-sided limits and the value.
If both one-sided limits are the same number but the value is missing or different, the discontinuity is removable. If both are numbers but different numbers, it is a jump. If either one grows without bound, it is infinite.
Removable: one point missing or out of place
Let . The numerator is (x − 2)(x + 1), so f(x) = x + 1 for every , and at x = 2 the expression reads and has no value. From both sides the values close in on 2 + 1 = 3. The graph is the line y = x + 1 with a hole at (2, 3).
Defining f(2) = 3 fills the hole, and the new function is continuous at 2. That is why this type is called removable: one value, chosen to be the limit, mends it.
A point that is present but in the wrong place is removable too. If f(x) = x + 1 for and f(2) = 1, the limit is 3 and the value is 1; changing the value at 2 to 3 mends it.
A removable discontinuity: the line y = x + 1 with a hollow dot at (2, 3). Both sides close in on 3, and the function has no value at 2.
Also removable: f(x) = x + 1 for and f(2) = 1. The hollow dot at (2, 3) is where both sides head; the solid dot at (2, 1) is the value. Moving the value to 3 mends the curve.
Jump: two different one-sided limits
Let f(x) = −1 for x < 0 and f(x) = 1 for . From the left the values are all −1, so the limit from the left is −1. From the right they are all 1, so the limit from the right is 1. The two one-sided limits are numbers, and different numbers, so there is no single limit at 0: a jump discontinuity, of size 1 − (−1) = 2.
No single point can mend a jump. Whatever value f(0) is given, it can match at most one side, because the two sides head for different heights. Here f(0) = 1 matches the right-hand side, so f is continuous from the right at 0 but not from the left.
A jump discontinuity: y = −1 for x < 0 and y = 1 for . The hollow dot at (0, −1) is where the left side ends without reaching; the solid dot at (0, 1) is the value f(0) = 1.
L⁻ = 1 but L⁺ = 4: the probes land 3 apart however close they come, so lim f(x) as x → 2 does not exist, even though each one-sided limit does
Bring both probes to x = 2, then close the jump
Two probes either side of x = 2. The left piece heads for 1 and the right piece for 4, so the probes land 3 apart however close they come. Moving one point cannot close that gap; shifting the whole right-hand piece down by 3 can.
Infinite: running off along an asymptote
Let . As from the right, 1 is divided by smaller and smaller positive numbers, and the values grow without bound: f(0.1) = 10, f(0.01) = 100. From the left they fall without bound: f(−0.01) = −100. The curve runs along the vertical line x = 0, its vertical asymptote.
Neither one-sided limit is a number, so there is no limit at 0, and no value given to f(0) could be one. This is an infinite discontinuity. It need not have opposite signs: grows without bound on both sides of 0, and that is an infinite discontinuity too.
An infinite discontinuity: , with its two branches either side of the dashed asymptote x = 0. On the right the curve climbs without bound as ; on the left it falls without bound.
at x = 2 the denominator is 0 and the numerator is not, so f grows without bound: a vertical asymptote, with opposite signs either side
Bring x to 0.1 from the forbidden point
Two functions with no value at x = 2. With , the denominator is 0 there and the numerator is not, and the values run off with opposite signs either side: infinite. With , both are 0, the expression equals x + 2 everywhere else, and both sides head for 4: removable.
Telling them apart
Find the limit from the left, the limit from the right, and the value. Equal limits with the value missing or different: removable. Two different numbers: jump. Either side without bound: infinite.
For a rational function, the denominator tells you where to look. Where a factor cancels with the numerator, the discontinuity is removable; where a factor of the denominator is left after canceling, it is infinite. For a function in pieces, look at the joins: a jump where the pieces arrive at different heights.
The usual mistakes
Calling a jump removable. A removable discontinuity needs both sides to agree; at a jump they differ, and no single value can mend it.
Calling an asymptote a jump. At a jump both sides settle on finite heights; beside an asymptote they do not settle at all.
Saying a jump function has no value at the jump. It usually has one, as f(0) = 1 above; what it lacks is a limit.
Drawing both ends of a jump as solid dots. The function takes one value at the point, so one end is hollow.
A flow gauge, a goods lift and a lens
The three applications below give one type each: a formula that reads at one sensor reading and is repaired by storing its limit; a floor indicator that steps up a whole floor at each landing; and a lens formula whose denominator is 0 at the focal length.
Worked example: A Flow Gauge That Shows an Error at Two Volts: The Value Stored to Repair It
Question A flow gauge turns a sensor reading of x volts into a flow of Q(x) = x3 − 8x − 2 liters per minute. At a reading of exactly 2 volts the gauge shows an error instead of a flow. (a) Find the value that should be stored for x = 2 so that Q is continuous there. (b) Find the flow at a reading of 2.5 volts.
1.Put x = 2 into the formula: the top is 23 − 8 = 0 and the bottom is 2 − 2 = 0. The form 00 is not a number, which is why the meter has nothing to show.
At x = 2 the top is 23 − 8 = 0 and the bottom is 2 − 2 = 0, so the formula asks for 00. 2.Factorize the top as a difference of two cubes: x3 − 8 = (x − 2)(x2 + 2x + 4).
Factorize the top as a difference of two cubes: x3 − 8 = (x − 2)(x2 + 2x + 4). 3.On the way to 2 the reading is near 2 and never equal to 2, so x − 2 is not zero and cancels: Q(x) = x2 + 2x + 4 for every x ≠ 2.
On the way to 2 the reading is never equal to 2, so x − 2 cancels and Q(x) = x2 + 2x + 4 for every x ≠ 2. 4.That polynomial is continuous everywhere, so limx → 2 Q(x) = 22 + 2 × 2 + 4 = 12. (a) Storing 12 liters per minute for x = 2 makes the reading continuous there, so the discontinuity is removable.
(a) That polynomial is continuous, so limx → 2 Q(x) = 12. Storing 12 liters a minute repairs the reading. 5.(b) At 2.5 volts, Q(2.5) = 2.52 + 2 × 2.5 + 4 = 6.25 + 5 + 4 = 15.25 liters per minute. Check against the original formula: 2.53 − 82.5 − 2 = 7.6250.5 = 15.25.
(b) At 2.5 volts the flow is 6.25 + 5 + 4 = 15.25 liters a minute, and the original formula agrees.
Answer: (a) 12 liters per minute; (b) 15.25 liters per minute
Common mistakes
- Reading 00 as 0, or as 1, and storing that. The form says only that the canceling has not been done yet; the number it hides here is 12.
- Writing Q(2) = 12 as though the original formula gave it. The original formula has no value at 2 at all; 12 is the value chosen for the point, and the repaired function is a new function that agrees with Q everywhere else.
Worked example: A Goods Lift's Floor Indicator: The Reading Either Side of a Floor
Question The floors of a warehouse are 3.5 m apart. A goods lift's indicator shows the floor number F(h) when the lift floor is h meters above the ground: 0 for 0 ≤ h < 3.5, then 1 for 3.5 ≤ h < 7, then 2 for 7 ≤ h < 10.5, and so on. (a) Classify the discontinuity of F at h = 7, giving both one-sided limits, the size of the jump, and the side from which F is continuous there. (b) Find the height at which the indicator first shows 4.
1.Just below 7 m the lift is still in the band from 3.5 m to 7 m, where the indicator reads 1, so limh → 7− F(h) = 1.
Just below 7 m the lift is still in the band from 3.5 m to 7 m, so limh → 7− F(h) = 1. 2.At 7 m and just above it the lift is in the next band, where the indicator reads 2, so F(7) = 2 and limh → 7+ F(h) = 2.
At 7 m and just above it the lift is in the next band, so F(7) = 2 and limh → 7+ F(h) = 2. 3.(a) Both one-sided limits are numbers but they differ, so this is a jump discontinuity, of size 2 − 1 = 1 floor. The value F(7) = 2 agrees with the limit from the right, so F is continuous from the right at 7 m and not from the left.
(a) The one-sided limits are numbers and they differ, so this is a jump discontinuity of size 1 floor; F(7) = 2, so F is continuous from the right. 4.For part (b), every band is 3.5 m tall and the ground band shows 0, so the band showing the floor number n begins at 3.5n meters.
Every band is 3.5 m tall and the ground band shows 0, so the band showing n begins at 3.5n meters. 5.(b) The indicator first shows 4 at 3.5 × 4 = 14 m. Check: at 13.9 m the reading is still 3, and it stays at 4 until 17.5 m.
(b) The indicator first shows 4 at 3.5 × 4 = 14 m, and at 13.9 m it still reads 3.
Answer: (a) a jump discontinuity: limh → 7− F(h) = 1 and limh → 7+ F(h) = 2, a jump of 1 floor, and F is continuous from the right at 7 m; (b) 14 m
Common mistakes
- Saying that F has no value at 7 m. It has the value 2; what it has no value for is the limit, because the two one-sided limits disagree.
- Counting the bands from 1 and answering 3.5 × 3 = 10.5 m. The ground band already shows 0, so the band showing 4 is the fifth band and begins at 3.5 × 4.
Worked example: A Lens at Its Focal Length: The Image Distance Near the Focal Point
Question A thin lens has a focal length of 20 cm. An object u cm from the lens gives an image v = 20uu − 20 cm from the lens, where a negative v means the image is on the same side of the lens as the object. (a) Classify the discontinuity of v at u = 20, and explain why no value given to v at u = 20 can make it continuous there. (b) Find the object distance that gives v = 100, an image 100 cm beyond the lens.
1.As u approaches 20 from above, the top tends to 20 × 20 = 400, a fixed positive number, and the bottom u − 20 tends to 0 through positive values.
As u → 20+ the top tends to 20 × 20 = 400, a fixed positive number, while the bottom u − 20 tends to 0 through positive values. From below it tends to 0 through negative values, and v falls without bound. 2.A fixed positive number divided by a smaller and smaller positive number grows without bound: at u = 20.1 the image is 4020.1 = 4020 cm, and at u = 20.001 it is 400.020.001 = 400020 cm.
So the image distance grows without bound: v(20.1) = 4020.1 = 4020 cm and v(20.001) = 400020 cm. 3.(a) So the image distance passes every length as u → 20+. From below, the bottom tends to 0 through negative values, so v falls without bound: at u = 19.9 it is 398−0.1 = −3980 cm. The discontinuity at u = 20 is an infinite discontinuity. Continuity at a point needs the limit there to be a number equal to the value; this limit is not a number at all, so no value given to v at u = 20 can repair it. An object sitting at the focal point makes no image.
(a) The break at u = 20 is an infinite discontinuity. Continuity needs a limit that is a number, and this limit is not one, so no value at u = 20 can repair it. 4.For part (b), set the image distance to 100: 20uu − 20 = 100. Multiply both sides by u − 20, which is not zero for u ≠ 20, to get 20u = 100u − 2000.
For part (b), 20uu − 20 = 100. Multiplying by u − 20, which is not zero for u ≠ 20, gives 20u = 100u − 2000. 5.(b) Take 20u from both sides: 80u = 2000, so u = 25 cm. Check: 20 × 2525 − 20 = 5005 = 100 cm, and v is positive, so the image is real and beyond the lens, as the question asks.
(b) 80u = 2000, so u = 25 cm, and the check gives 5005 = 100 cm.
Answer: (a) an infinite discontinuity: the top tends to 400 while the bottom tends to 0, so v grows without bound on each side and no value at u = 20 can be its limit; (b) u = 25 cm
Common mistakes
- Calling the break at u = 20 removable and writing some large value there. A removable discontinuity has a limit that is a number; here the one-sided limit is not a number, so no value works.
- Multiplying through by u − 20 without saying why it is not zero. The working holds only for u ≠ 20, which is exactly the point where the formula has no value.