The limit laws
Reading a limit off a graph or a table works, but it is slow and never quite certain. If you already know where two functions are heading, can you say exactly where their sum, product, quotient or composition is heading, without drawing anything?
Where you are 11 in Limits
Why this exists
In What a limit is: graphs and tables a limit was read by sliding along a curve toward x = a from both sides, or estimated from a table of inputs closing in on a. Both methods are evidence, not proof, and both are slow. Yet most functions in calculus are built from a few simple pieces by adding, multiplying, dividing and composing. If the limit of each piece is known, the limit laws give the limit of the whole by doing the same arithmetic on the pieces’ limits. They turn a limit from something estimated into something computed exactly, and they show that for polynomials, and for many other functions, the limit at a is simply the value at a.
The lesson takes two days. Part 1 is the sum, difference, constant multiple and product laws, and what they say about polynomials; Examples 1 to 3 go with it. Part 2 is the quotient and root laws, direct substitution, and the limit of a composition; Examples 4 to 7 go with it.
Definition
In plain words: when two functions each have a limit at a, the limit of their sum, difference, product or quotient is the sum, difference, product or quotient of their limits, with one exception, a quotient whose bottom limit is 0.
In symbols, let a be a number and suppose
where A and B are numbers, so both limits exist. Then
Here k is any constant, and the last line holds only when B \ne 0. The hypothesis “both limits exist” is part of every law. The laws say nothing about f(a) or g(a): like the limit itself, they look only at inputs near a, never at a.
Intuition
A limit is a promise about closeness: once x is near enough to a, f(x) is as close to A as you like. If f(x) is within 0.01 of A and g(x) is within 0.01 of B, then f(x) + g(x) is within 0.02 of A + B, since the two errors can at worst add up. Shrink the errors in f and g and the error in the sum shrinks with them (Figure 1). The same is true of a product, because a small change in either factor makes only a small change in the product, and of a quotient as long as the bottom stays away from 0. Dividing by a number near 0 magnifies errors without limit, which is why the quotient law needs B \ne 0.
Composition works the same way. If g(x) is close to L and f is a function that changes by only a little when its input changes by a little near L, then f(g(x)) is close to f(L). A function with that property at L is continuous at L: its limit there equals its value, \lim_{u \to L} f(u) = f(L). This is the definition of Continuity, studied in Continuity at a point.
Results
Part 1: sums, differences, products and polynomials
Two basic limits. For any constant c and any number a,
The sum, difference, constant multiple and product laws are the first four lines of the definition. Applied again and again, the product law gives the power law: if \lim_{x \to a} f(x) = A and n is a positive integer, then \lim_{x \to a} \big[f(x)\big]^n = A^n.
Polynomials. For every polynomial p and every number a,
Part 2: quotients, roots, compositions
The quotient law is the last line of the definition, with its condition B \ne 0.
The root law. If \lim_{x \to a} f(x) = A and n is a positive integer, then \displaystyle \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{A}, provided A > 0 when n is even.
Direct substitution. If f is a polynomial, or a rational function \displaystyle \frac{p(x)}{q(x)} with q(a) \ne 0, then \lim_{x \to a} f(x) = f(a). The same holds for \displaystyle \sqrt[n]{x} at every a in its domain (with a > 0 when n is even), and for \sin x, \cos x and b^x (b > 0) at every a; these last are stated here without proof. A function for which \lim_{x \to a} f(x) = f(a) is exactly a function continuous at a.
The limit of a composition. If \lim_{x \to a} g(x) = L and f is continuous at L, then
Composition, feeding the output of g into f, is Combining and composing functions.
Other kinds of limit. Every law above holds word for word for the one-sided limits x \to a^- and x \to a^+ of One sided limits and when a limit exists, and for x \to \infty and x \to -\infty, as long as the limits involved are numbers.
Why it is true
The proofs rest on one fact about distances, the triangle inequality |u + v| \le |u| + |v|, and on reading “as close as you like” as “within any tolerance t > 0 that is named”. The exact version of this reading is The epsilon delta definition of a limit.
The basic limits. The outputs of a constant function are all equal to c, so they are within every tolerance of c. For f(x) = x, the distance from the output to a is the distance from the input to a, so the output is within t of a whenever the input is.
The sum law. Name a tolerance t > 0. Since f(x) \to A, there is a window around a in which every x \ne a has \displaystyle |f(x) - A| < \frac{t}{2}. There is also a window in which \displaystyle |g(x) - B| < \frac{t}{2}. In the smaller of the two windows both hold, and
So the sum is within t of A + B near a, and t was any tolerance.
The constant multiple law. |k f(x) - kA| = |k|\,|f(x) - A|. Near enough to a, \displaystyle |f(x) - A| < \frac{t}{|k| + 1}, and then |k f(x) - kA| < t. The difference law follows, since f - g = f + (-1)g.
The product law. Multiplying out the right side shows
Each of the three terms on the right has a factor that goes to 0, so by the sum and constant multiple laws (and the fact that a product of two quantities each near 0 is near 0), the whole right side goes to 0. That is, f(x)g(x) \to AB.
Polynomials. Write p(x) = c_n x^n + \cdots + c_1 x + c_0. By \lim x = a and the power law, x^k \to a^k; by the constant multiple law, c_k x^k \to c_k a^k; by the sum law, p(x) \to c_n a^n + \cdots + c_1 a + c_0 = p(a).
The quotient law. Since \displaystyle f/g = f \cdot \frac{1}{g}, it is enough to show \displaystyle \frac{1}{g(x)} \to \frac{1}{B}. Near a, g(x) is within \displaystyle \frac{|B|}{2} of B, so \displaystyle |g(x)| > \frac{|B|}{2}, which keeps g(x) away from 0. Then
and the right side goes to 0. Here is where B \ne 0 is used.
Composition. Name a tolerance t. Because f is continuous at L, there is a distance s such that |f(u) - f(L)| < t for every u within s of L. Because g(x) \to L, every x \ne a near enough to a has g(x) within s of L, so f(g(x)) is within t of f(L). The root law is this argument with \displaystyle f(u) = \sqrt[n]{u}, once that function is known to be continuous for u > 0, which is stated here without proof.
Worked examples
A polynomial, one law at a time
Find \lim_{x \to 3} (2x^2 - 5x + 1), naming the law behind each step.
SolutionThe limits \lim_{x \to 3} 2x^2, \lim_{x \to 3} 5x and \lim_{x \to 3} 1 all exist, as the steps below show, so the sum and difference laws split the limit into three:
The second line uses the constant multiple law twice, the power law on x^2, and the limit of a constant. The third uses \lim_{x \to 3} x = 3. So \lim_{x \to 3} (2x^2 - 5x + 1) = 4.
The answer is the value of the polynomial at 3, as the polynomial result promises. A table agrees: at x = 2.99 the polynomial is 3.9302 and at x = 3.01 it is 4.0702, on either side of 4.
Maiwenn’s limits with no formulas
Maiwenn knows only that \lim_{x \to 2} f(x) = 4 and \lim_{x \to 2} g(x) = -3. Find \lim_{x \to 2} \big[3f(x) - 2g(x)\big], \lim_{x \to 2} \big[x f(x) + \big(g(x)\big)^2\big] and \lim_{x \to 2} \big[f(x) - g(x)\big]^2.
SolutionThe laws need only the two limits, not formulas. By the constant multiple and difference laws,
For the second, \lim_{x \to 2} x = 2, so the product law gives x f(x) \to 2 \cdot 4 = 8; the power law gives \big(g(x)\big)^2 \to (-3)^2 = 9; and the sum law adds them:
For the third, the difference law gives f(x) - g(x) \to 4 - (-3) = 7, and the power law squares it:
Nothing was said about f(2) or g(2); the functions might not even be defined at 2. The answers stand either way.
Gwenael’s product with a moved point
Let p(x) = 3 - x for x \ne 1, and p(1) = 6. Gwenael wants \lim_{x \to 1} (x^2 + 3)\,p(x), and also the value of (x^2 + 3)\,p(x) at x = 1.
SolutionThe graph of p is the line y = 3 - x with the point above 1 moved to height 6. The limit looks only at x \ne 1, where p(x) = 3 - x, so \lim_{x \to 1} p(x) = 3 - 1 = 2. The polynomial x^2 + 3 has limit 1 + 3 = 4. Both limits exist, so the product law gives
The value at 1 is a different question: (1^2 + 3) \cdot p(1) = 4 \cdot 6 = 24. Plugging x = 1 into the whole expression would have given 24, which is the value, not the limit. The product law combines limits, and the limit of p at 1 is 2, not p(1). Direct substitution fails here because p is not continuous at 1.
A quotient whose bottom limit is not zero
Find \lim_{x \to -1} \dfrac{x^3 + 4}{x^2 + 2}.
SolutionBefore using the quotient law, find the two limits separately and confirm the bottom one is not 0. Both top and bottom are polynomials, so their limits are their values:
The bottom limit is 3 \ne 0, so the quotient law applies:
This is direct substitution for a rational function, allowed because the bottom is not 0 at -1. In fact x^2 + 2 \ge 2 for every x, so this function is continuous everywhere.
The limit of a square root
Find \displaystyle \lim_{x \to 3} \sqrt{x^2 + 7}.
SolutionThis is a composition: the inner function g(x) = x^2 + 7 feeds the outer function \displaystyle f(u) = \sqrt{u}. The inner limit is
The square root is continuous at 16, since 16 > 0. So the limit of a composition (here the root law with n = 2) gives
The limit moved inside the square root, and the root was taken last. Since x^2 + 7 > 0 for every x, the same reasoning gives \displaystyle \lim_{x \to a} \sqrt{x^2 + 7} = \sqrt{a^2 + 7} at every a.
Choosing a constant to fix a limit
Find every constant k for which \lim_{x \to 2} \dfrac{k x^2 + 3}{x + k} = 3.
SolutionThe top tends to 4k + 3 and the bottom to 2 + k. If k = -2, the bottom tends to 0 while the top tends to 4(-2) + 3 = -5; dividing -5 by numbers closer and closer to 0 gives outputs that grow without bound, so the limit does not exist and k = -2 fails. For every other k, the bottom limit is not 0 and the quotient law gives
Setting this equal to 3 and multiplying by 2 + k, which is not 0:
Check: with k = 3 the limit is \displaystyle \frac{12 + 3}{2 + 3} = \frac{15}{5} = 3. So k = 3 is the only such constant.
Loig’s composition through a jump
Let f(u) = 1 for u < 2 and f(u) = 3 for u \ge 2 (Figure 2), and let g(x) = 2 - x^2. Loig finds \lim_{x \to 0} g(x) = 2 and f(2) = 3, and concludes that \lim_{x \to 0} f\big(g(x)\big) = 3. Is he right?
His first two facts are correct: g is a polynomial, so \lim_{x \to 0} g(x) = g(0) = 2, and f(2) = 3 from the second rule. But the composition law needs f to be continuous at L = 2, and f jumps there, so the law cannot be used.
Look at the actual outputs instead. For every x \ne 0, x^2 > 0, so g(x) = 2 - x^2 < 2. Then f\big(g(x)\big) uses the first rule and equals 1. Near 0, except at 0 itself, the composition is the constant 1, so
The value at 0 is f(g(0)) = f(2) = 3, which is what Loig computed: he found the value of the composition, not its limit. The inner function approaches 2 only from below, so only the left side of the jump in f matters. Had g(x) been 2 + x^2, approaching from above, the limit would have been 3.
Boundaries
- Both limits must exist. The laws turn two existing limits into a third. They cannot be run backward to split a limit that exists into pieces that do not. With \displaystyle f(x) = \frac{1}{x} and \displaystyle g(x) = 3 - \frac{1}{x}, neither \lim_{x \to 0} f(x) nor \lim_{x \to 0} g(x) exists, since the outputs grow without bound, yet f(x) + g(x) = 3 for every x \ne 0, so \lim_{x \to 0} \big[f(x) + g(x)\big] = 3. Writing it as “\lim f + \lim g” would add two things that are not numbers.
- The quotient law needs the bottom limit to be nonzero. For \displaystyle \lim_{x \to 2} \frac{x + 1}{x - 2} the top tends to 3 and the bottom to 0; the outputs grow without bound and the limit does not exist. When top and bottom both tend to 0, the form \displaystyle \frac{0}{0} says nothing at all: such a limit may be any number or may not exist, and finding it takes the algebra of Limits by algebra: factoring and rationalizing.
- An even root needs a positive limit underneath. If f(x) \to A < 0, then \displaystyle \sqrt{f(x)} is undefined near a. If A = 0, the root may be defined on one side only, as \displaystyle \sqrt{x} is at 0, and only a one-sided limit can be asked.
- Direct substitution needs continuity. The limit of a function at a equals its value only when the function is continuous at a. Example 3 shows a product whose limit (8) and value (24) differ.
- The composition law needs the outer function continuous at the inner limit. Example 7 shows a composition whose limit is not f(L) when f jumps at L.
- Infinity is not a number. A statement like \lim_{x \to a} f(x) = \infty records how a limit fails to exist, so none of the laws applies to it.
Where this goes next
Direct substitution is the bridge to Continuity at a point: the laws show at once that sums, products, quotients (bottom nonzero) and compositions of continuous functions are continuous. When substitution gives \displaystyle \frac{0}{0}, Limits by algebra: factoring and rationalizing rewrites the function so that the laws apply again. Limits the laws and algebra cannot reach, such as \displaystyle \lim_{x \to 0} \frac{\sin x}{x}, need the squeeze theorem. The laws hold as x \to \infty too, which is how Limits at infinity and horizontal asymptotes finds the level a rational function settles to. Most of all, the derivative is a limit, and the sum and constant multiple laws for limits become the sum and constant multiple rules of differentiation.
Exercises
Part 1: polynomials and products
- Find \lim_{x \to 2} (x^3 - 4x + 7), naming the law behind each step.
- Find \lim_{x \to -3} (4x^2 + 5x - 2).
- Find \lim_{x \to 1/2} (6x^2 - x + 3).
- Find \lim_{x \to -1} (2x + 5)(x^2 - 3x) by the product law.
- Find \lim_{x \to 4} (x - 1)^3 by the power law.
Part 1: from known limits
For Exercises 6 to 8, \lim_{x \to 5} f(x) = -2 and \lim_{x \to 5} g(x) = 6.
- Find \lim_{x \to 5} \big[4f(x) + g(x)\big].
- Find \lim_{x \to 5} \big[f(x)\,g(x) - x\big].
- Find \lim_{x \to 5} \big[\big(f(x)\big)^3 + \big(g(x)\big)^2\big].
- Let q(x) = 2x - 1 for x \ne 3 and q(3) = 0. Find \lim_{x \to 3} \big[x^2 + q(x)\big] and the value of x^2 + q(x) at x = 3.
Part 2: quotients, roots and compositions
- Find \lim_{x \to 2} \dfrac{x^2 + 5}{3x - 1}.
- Find \lim_{x \to -2} \dfrac{x^3 + 1}{x^2 - x}.
- Find \displaystyle \lim_{x \to -1} \sqrt[3]{x^2 - 9}.
- Find \displaystyle \lim_{x \to 4} \sqrt{x^2 + 9}.
- Find \lim_{x \to 2} 2^{x^2 - 1}, using that 2^u is continuous everywhere.
Part 2: known limits and where the laws stop
- Given \lim_{x \to 1} f(x) = 9 and \lim_{x \to 1} g(x) = -4, find \displaystyle \lim_{x \to 1} \dfrac{\sqrt{f(x)} - g(x)}{f(x) + g(x)}.
- Find every constant k for which \lim_{x \to 1} \dfrac{x^2 + k}{kx + 2} = 2.
- Does the quotient law give \lim_{x \to 3} \dfrac{x + 2}{x^2 - 9}? Describe the outputs on each side of 3 and decide whether the limit exists.
- Let f(u) = 2 for u < 1 and f(u) = 5 for u \ge 1, and let g(x) = 1 + x. Find \lim_{x \to 0^-} f\big(g(x)\big) and \lim_{x \to 0^+} f\big(g(x)\big). Does \lim_{x \to 0} f\big(g(x)\big) exist?
Reasoning
- Give functions f and g such that neither \lim_{x \to 0} f(x) nor \lim_{x \to 0} g(x) exists, but \lim_{x \to 0} f(x)\,g(x) does.
- Explain how the difference law follows from the sum law and the constant multiple law.
- Use the laws to explain why \lim_{x \to a} p(x) = p(a) for the polynomial p(x) = 7x^4 - 2x + 9 at every number a.
- Maiwenn says: “If \lim_{x \to a} \big[f(x) + g(x)\big] and \lim_{x \to a} f(x) both exist, then \lim_{x \to a} g(x) exists.” Is she right? Explain.
- Explain why the quotient law needs B \ne 0, using one example where the top limit is not 0 and two examples where both limits are 0 but the limits of the quotients differ.
Answers
- 7. Sum and difference laws split it into \lim x^3 - \lim 4x + \lim 7; the power, constant multiple and constant laws give 2^3 - 4 \cdot 2 + 7 = 8 - 8 + 7 = 7.
- 19. It is a polynomial, so the limit is its value: 4 \cdot 9 - 15 - 2 = 19.
- 4. The value at \displaystyle \frac{1}{2}: \displaystyle 6 \cdot \frac{1}{4} - \frac{1}{2} + 3 = \frac{3}{2} - \frac{1}{2} + 3 = 4.
- 12. The factors tend to 2(-1) + 5 = 3 and 1 + 3 = 4, and 3 \cdot 4 = 12.
- 27. x - 1 \to 3, and the power law gives 3^3 = 27.
- -2. 4(-2) + 6 = -2.
- -17. f(x)\,g(x) \to (-2)(6) = -12 and x \to 5, so the limit is -12 - 5 = -17.
- 28. (-2)^3 + 6^2 = -8 + 36 = 28.
- The limit is 14 and the value is 9. Near 3 (not at it) q(x) = 2x - 1 \to 5, so x^2 + q(x) \to 9 + 5 = 14; at 3 the value is 9 + q(3) = 9 + 0 = 9.
- \displaystyle \frac{9}{5}. Top \to 9, bottom \to 5 \ne 0.
- \displaystyle -\frac{7}{6}. Top \to -8 + 1 = -7, bottom \to 4 + 2 = 6 \ne 0.
- -2. x^2 - 9 \to -8, and a cube root needs no sign condition: \displaystyle \sqrt[3]{-8} = -2.
- 5. x^2 + 9 \to 25 > 0, and \displaystyle \sqrt{25} = 5.
- 8. x^2 - 1 \to 3, and 2^u is continuous at 3, so the limit is 2^3 = 8.
- \displaystyle \frac{7}{5}. \displaystyle \sqrt{f(x)} \to \sqrt{9} = 3 (root law, 9 > 0), so the top \to 3 - (-4) = 7; the bottom \to 9 + (-4) = 5 \ne 0.
- k = -3. If k = -2 the bottom tends to 0 while the top tends to -1, so there is no limit. Otherwise the limit is \displaystyle \frac{1 + k}{k + 2} = 2, so 1 + k = 2k + 4 and k = -3. Check: \displaystyle \frac{1 - 3}{-3 + 2} = \frac{-2}{-1} = 2.
- No: the bottom tends to 0 while the top tends to 5. Just left of 3 the bottom is a small negative number, so the outputs fall without bound (\to -\infty); just right of 3 they rise without bound (\to \infty). The limit does not exist.
- 2 and 5; the limit does not exist. For x < 0, g(x) < 1 and f(g(x)) = 2; for x > 0, g(x) > 1 and f(g(x)) = 5. The one-sided limits differ.
- One example: f(x) = g(x) = 1 for x \ge 0 and -1 for x < 0. Each jumps at 0 and has no limit there, but f(x)\,g(x) = 1 for every x, so its limit is 1.
- f(x) - g(x) = f(x) + (-1)\,g(x). The constant multiple law gives (-1)\,g(x) \to -B, and the sum law gives f(x) + (-1)\,g(x) \to A + (-B) = A - B.
- \lim x = a, so the power law gives x^4 \to a^4; the constant multiple law gives 7x^4 \to 7a^4 and -2x \to -2a; the constant 9 has limit 9; the sum law adds them to 7a^4 - 2a + 9 = p(a).
- Yes. g(x) = \big[f(x) + g(x)\big] - f(x) is a difference of two functions whose limits exist, so by the difference law \lim g(x) exists and equals \lim \big[f(x) + g(x)\big] - \lim f(x).
- With B = 0 and A \ne 0, dividing by numbers near 0 gives outputs that grow without bound: \displaystyle \frac{1}{x} has no limit at 0. With A = B = 0, the answer depends on the functions: \displaystyle \frac{x}{x} = 1 for x \ne 0 has limit 1, while \displaystyle \frac{x^2}{x} = x for x \ne 0 has limit 0. Since “\displaystyle \frac{A}{B}” cannot give an answer in either case, the law has to exclude B = 0.