Without lifting the pencil
Informally, a function is continuous at a point if its graph can be drawn through that point without lifting the pencil: no gap, no step, no point out of place.
Take at x = 2. The value is f(2) = 4 − 1 = 3. Just either side of 2 the values are close to 3: f(1.99) = 2.9601 and f(2.01) = 3.0401. Nothing breaks there.
The limit equals the value
Formally, f is continuous at x = a when lim f(x) as equals f(a). The limit says where the graph is heading as x closes in on a from both sides; f(a) says where the graph actually is at a. Continuity is the two agreeing: both sides arrive at the point the function already takes.
For the line y = x + 2 at a = 2, the limit is 4 and the value is 4, so it is continuous there.
y = x + 2, with a solid dot at (2, 4). From both sides the curve heads for height 4, and at x = 2 the function takes the value 4.
Three conditions
lim f(x) = f(a) packs three conditions into one equation, and all three must hold. First, f(a) exists: the function has a value at a. Second, the limit as exists: the values from the left and from the right close in on one number. Third, the two are equal.
Any one failing breaks the graph at a. The three pictures below keep the line y = x + 2 near a = 2 and break one condition each.
No value at 2: , which equals x + 2 except at x = 2, where it reads . The limit is 4, but there is no f(2) for it to match. The hollow dot marks the missing point.
A value, but not the limit: f(x) = x + 2 for , and f(2) = 7. The curve heads for 4 from both sides, and the solid dot at (2, 7) is where the function is. The limit and the value both exist, and they differ.
No limit: f(x) = x + 2 for x < 2 and x + 4 for . From the left the values close in on 4, from the right on 6, so there is no single limit at 2, though f(2) = 6 exists. The hollow dot at (2, 4) is not on the graph; the solid dot at (2, 6) is.
Functions that are continuous everywhere
For a polynomial, the limit laws give the limit at any a by substituting a, which is exactly f(a). So every polynomial is continuous at every point. Sine and cosine are continuous everywhere too.
A rational function, one polynomial over another, is continuous at every a where its denominator is not 0, since there the limit of the quotient is the quotient of the values. Where the denominator is 0 the function has no value, so it is not continuous there.
A function is continuous on an interval when it is continuous at every point of it.
The join of a piecewise function
A function given in pieces is continuous inside each piece if each rule is, so the joins are what need testing. At a join, take the limit from the left with the rule on the left, the limit from the right with the rule on the right, and the value with whichever rule includes the point.
Let g(x) = 2x for x < 1 and g(x) = x + k for . From the left, g tends to 2 × 1 = 2. From the right, and at the point itself, the rule gives 1 + k. Continuity at 1 needs 1 + k = 2, so k = 1. Check: with k = 1 both sides and the value are 2.
The usual mistakes
Taking "the limit exists and the value exists" as enough. A limit of 4 with f(2) = 7 has both, and the function is not continuous at 2: they must also be equal.
Letting the limit fill the hole. If the limit at 1 is 3 but f(1) has no value, f is not continuous at 1. The limit describes the approach; it does not supply the missing value.
Reading f(0) = 0 as no value. 0 is a value, so if the limit at 0 is also 0, the function is continuous there.
Testing only one join of a piecewise function. Every join has its own pair of one-sided limits, and one join holding says nothing about the next.
An electricity tariff and an oven program
In the first application below, a bill is charged by one rule up to 200 units and by another above, and the rate above is chosen so that the two rules meet at the join. In the second, an oven program in three pieces is tested at each of its two joins.
Worked example: A Two-Band Electricity Tariff: The Rate Above the Band That Leaves No Jump
Question A power company charges $0.28 for each unit of electricity used, up to 200 units in a month. When more than 200 units are used the whole bill is C(x) = ax + 12 dollars, where x is the number of units used. (a) Find the value of a for which the bill is continuous at x = 200. (b) Using that value of a, find the bill for a month in which 350 units are used.
1.Below the join the first rule applies, so C(x) = 0.28x for 0 ≤ x ≤ 200. The limit from the left is limx → 200− C(x) = 0.28 × 200 = 56, that is $56.
Below the join the first rule holds, so limx → 200− C(x) = 0.28 × 200 = 56 dollars. 2.Above the join the second rule applies, so the limit from the right is limx → 200+ C(x) = a × 200 + 12 = 200a + 12 dollars.
Above the join the second rule holds, so limx → 200+ C(x) = 200a + 12 dollars. 3.Continuity at 200 units means the two one-sided limits agree with each other and with the value there, so 200a + 12 = 56.
Continuity at 200 units means the two one-sided limits are the same number: 200a + 12 = 56. 4.Subtract 12 from both sides to get 200a = 44, then divide both sides by 200. (a) a = 0.22, so the rate above the band is $0.22 for each unit.
(a) a = 0.22. The second piece now starts at the value the first one reaches, so the bill has no break at 200 units. 5.(b) A month of 350 units uses the second rule: C(350) = 0.22 × 350 + 12 = 77 + 12 = $89. Check: at 200 units the second rule gives 0.22 × 200 + 12 = $56, the same as the first rule, so the graph has no break.
(b) A month of 350 units costs 0.22 × 350 + 12 = $89.
Answer: (a) a = 0.22; (b) $89
Common mistakes
- Setting a × 200 = 56 and leaving the standing charge of $12 out of the equation. Continuity compares the whole bill on each side of the join, not only the part of it that varies with the units.
- Charging the second rate on the units above 200 only, as a stepped tariff does. The second rule here is written for the whole bill, so its x is the total number of units.
Worked example: An Oven Program in Three Parts: One Join That Holds and One That Breaks
Question An oven program sets the temperature T(m) degrees Celsius m minutes after it starts: T(m) = 20 + 8m for 0 ≤ m ≤ 10, T(m) = 60 + 4m for 10 < m ≤ 25, and T(m) = 200 − m for m > 25. (a) Show that the program is continuous at m = 10. (b) Show that it is not continuous at m = 25, find the size of the jump, and find the constant that must replace 200 to remove it.
1.At the first join the left rule gives the value and the limit from the left: T(10) = 20 + 8 × 10 = 100, so limm → 10− T(m) = 100.
The left rule gives both the value and the limit from the left at the first join: T(10) = 20 + 8 × 10 = 100. 2.The middle rule gives limm → 10+ T(m) = 60 + 4 × 10 = 100. (a) The value and both one-sided limits are 100, so the program is continuous at m = 10.
(a) The middle rule gives limm → 10+ T(m) = 60 + 4 × 10 = 100 as well, so the program is continuous at m = 10. 3.At the second join the middle rule gives limm → 25− T(m) = 60 + 4 × 25 = 160, and the last rule gives limm → 25+ T(m) = 200 − 25 = 175.
At the second join the middle rule gives limm → 25− T(m) = 160 and the last rule gives limm → 25+ T(m) = 175. 4.(b) The two one-sided limits are different numbers, so the program is not continuous at m = 25: it jumps by 175 − 160 = 15 degrees Celsius.
(b) The two one-sided limits differ, so the program is not continuous at m = 25: it jumps by 175 − 160 = 15 degrees Celsius. 5.Write the last rule as c − m and ask for no jump at m = 25: c − 25 = 160, so c = 185. Replacing 200 by 185 removes the jump. Check: 185 − 25 = 160, which is the temperature the middle rule reaches.
Writing the last rule as c − m and asking for c − 25 = 160 gives c = 185, which closes the join.
Answer: (a) continuous at m = 10, where both pieces give 100 degrees Celsius; (b) not continuous at m = 25: the limit from the left is 160 and the limit from the right is 175, a jump of 15 degrees Celsius, and replacing 200 by 185 removes it
Common mistakes
- Testing the first join, finding it sound, and declaring the whole program continuous. A rule in pieces has to be tested at every join, and here one join holds while the other does not.
- Taking the jump as 200 − 160 = 40. The jump is the difference of the two one-sided limits at m = 25, and the limit from the right is 200 − 25 = 175, not 200.