Substituting gives 0 over 0
The first move with any limit is to substitute. For as , putting x = 2 in gives a numerator of 4 − 4 = 0 and a denominator of 2 − 2 = 0, so the substitution produces .
is not a number. A quotient would have to be a number c with c × 0 = 0, and every number does that, so no single value is picked out. The form is called indeterminate: it says that this expression, as written, cannot decide the limit, and that it has to be rewritten first. It does not say that the limit is missing.
Values close to 2 show that there is a limit to find. At x = 1.9 the expression is (3.61 − 4) ÷ (1.9 − 2) = −0.39 ÷ (−0.1) = 3.9; at 1.99 it is 3.99; at 2.01 it is 4.01; and at 2.1 it is 4.1. From both sides the values close in on 4.
(x² − 4)/(x − 2) = x + 2 everywhere except x = 2, so both sides approach 4 and only the single point is missing: a removable hole
Bring x to 0.1 from the forbidden point
The graph of : the line y = x + 2 with a hollow dot at (2, 4), where the expression has no value. Drag x toward 2 from either side: at 1.9 the value is 3.9 and at 2.1 it is 4.1. The other button draws , where the numerator is not 0 at x = 2, and there the values run off instead.
Factor and cancel
The numerator is a difference of two squares: . So .
Canceling divides the numerator and the denominator by x − 2, which is allowed whenever x − 2 is not 0. A limit as looks only at x near 2, never at x = 2 itself, so on the way to 2 the factor x − 2 is never 0 and the cancel is valid: for every .
The canceled function x + 2 equals the original everywhere except at x = 2. There the original has no value, since it reads , while x + 2 gives 4. The two graphs are the same line, except that the original has a hole at (2, 4).
Substitute again
Two functions that agree at every x near 2 have the same limit at 2, so as equals lim (x + 2) as . The second is a polynomial, and substituting into it is safe: 2 + 2 = 4.
So the limit is 4. It is the height of the hole: the value the expression closes in on from both sides, though the expression itself never takes it.
The method
Substitute. If the result is a number, that number is the limit. If it is , rewrite the expression so that the factor making both parts 0 cancels, and then substitute into the new expression.
Three rewrites cover most cases: factor and cancel, rationalize a surd, or combine fractions into one. Each uses only that x is near the point and not at it.
A nonzero number over 0 is a different signal. at x = 2 gives : the numerator is 1 while the denominator shrinks toward 0, so the values grow without bound, and there is no rewriting that gives a finite limit.
More factorizations
For as , substitution gives . Factor: for , and substituting gives 3 + 3 = 6. At x = 3.01 the original is 6.01.
For as , the numerator is a difference of two cubes: . For the expression is , which gives 4 + 4 + 4 = 12. At x = 1.9 the original is 11.41 and at 2.1 it is 12.61.
Both parts may need factoring. For as , the numerator is (x + 3)(x − 2) and the denominator is (x + 2)(x − 2). For the expression is , which gives . At x = 2.01 the original is 1.2494.
Rationalizing a surd
For as , substitution gives . The surd blocks any factoring, so multiply the numerator and the denominator by , the conjugate of .
The numerator becomes a difference of two squares: . So for the expression is , and substituting x = 9 gives .
Check: at x = 9.01 the original is 0.16662, and at x = 8.99 it is 0.16671. Both are close to .
The graph of for x from 0 to 20. It is the curve , falling from at x = 0, with a hollow dot at , where the original expression has no value.
Combining fractions
For as , substitution gives . Write the numerator as one fraction: .
Then the whole expression is . Since 2 − x = −(x − 2), the factor x − 2 cancels for and leaves . Substituting x = 2 gives . Check: at x = 2.01 the original is −0.24876, and at x = 1.99 it is −0.25126.
The usual mistakes
Reading as 0, because the numerator is 0, or as 1, because the numerator and the denominator are equal. Neither follows. For the limit is 4, and for it is 6, though both start as .
Taking the limit of one part. does not approach 9, the limit of alone: the whole fraction cancels to x + 3, which approaches 6.
Writing f(2) = 4 for . The limit is 4; the value at 2 does not exist. x + 2 is a second function that agrees with f everywhere except at 2.
Canceling terms rather than factors. In , striking out an x from the top and the bottom is not allowed. Only a factor of the whole numerator and the whole denominator may be canceled.
A spreadsheet that gives no answer, and a dish
In the first application below, a savings formula has r − 1 in its denominator. At r = 1 it reads , and the factor r − 1 cancels from , as multiplying out the right side shows: every middle term cancels.
In the second, the depth of a dish involves a surd, and the is cleared by multiplying by the conjugate, as for above.
Worked example: A Savings Club's Spreadsheet That Gives No Answer at No Growth: The Value the Formula Closes On
Question A savings club pays $200 into a fund at the end of each year for 20 years. The fund multiplies by r each year, so the value at the end is T(r) = 200 × r20 − 1r − 1 dollars. The club's spreadsheet reports an error for a year of no growth at all, which is r = 1. (a) Find limr → 1 T(r) and say what it means for the club. (b) With no growth at all, how many yearly payments of $200 would the club need to hold $6000?
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.
At r = 1 the top is 120 − 1 = 0 and the bottom is 1 − 1 = 0, so the formula asks for 00. 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.
Either side of 1 the formula does have a value, and those values are closing in on $4000. 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).
The top factorizes as (r − 1)(r19 + … + 1), and r − 1 is not zero on the way to 1, so it cancels. 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 limr → 1 T(r) = 200 × 20 = 4000.
What is left is 200(1 + r + … + r19). Each of the twenty terms tends to 1, so the sum tends to 20. 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.
(a) limr → 1 T(r) = 200 × 20 = $4000: the twenty payments with no growth added to any of them. 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.
(b) With no growth, 200n = 6000, so the club would need n = 30 payments.
Answer: (a) limr → 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
Common mistakes
- Reading 00 as 1, because the top and the bottom are equal, or as 0, because the top is zero. Neither follows. An indeterminate form takes its value from the way the top and the bottom approach zero, which is what the factorizing and canceling find out; here the answer is 4000, and neither guess is near it.
- Canceling r − 1 and then writing T(1) = 4000. The original formula has no value at r = 1, because dividing by zero is undefined. What is true is that T(r) tends to 4000 as r tends to 1: the canceled sum is a second function, one that agrees with T everywhere except at r = 1, where it fills the hole.
Worked example: The Depth of a Solar Cooker's Dish: A Ratio That Needs the Surd Rationalized
Question The reflector of a solar cooker is part of a circle of radius 2 m. At a horizontal distance a meters from the axis the dish is s(a) = 2 − √4 − a2 meters deep. Makers of shallow dishes use the rule that the depth is a24. (a) Find lima → 0 s(a)a2, and say what it has to do with that rule. (b) The dish is 1.2 m across, so a = 0.6 m at the rim. Compare the rule's depth there with the exact depth.
1.Substituting a = 0 gives 2 − √40 = 00, an indeterminate form. The surd is what blocks any canceling, so clear it first.
At a = 0 the ratio is 00, an indeterminate form, and the surd is what blocks any canceling. 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.
Multiply the top and the bottom by the conjugate 2 + √4 − a2: the top becomes 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.
The a2 that made the bottom zero cancels, leaving 12 + √4 − a2. 4.Substituting is safe now: 12 + √4 − 0 = 14.
Substituting is safe now: 12 + √4 = 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.
(a) The limit is 14, that is 0.25 per meter, which is one over twice the radius and is where the rule comes from. 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.
(b) At the rim the rule gives 0.364 = 0.09 m against an exact 2 − √3.64 = 0.0921 m, about 2 mm shallow.
Answer: (a) lima → 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
Common mistakes
- Writing √4 − a2 = 2 − a. The root of a difference is not the difference of the roots: at a = 0.6, √3.64 = 1.908, while 2 − 0.6 = 1.4, and the depth would come out ten times too big.
- Reading the limit 14 as a depth. It is a depth divided by the square of a distance, so it is measured per meter; the depth itself is about a24 meters, and that tends to 0 as a does.