Source: Homework 03, Section 1.6 – Limit Laws | Calculus I, Michigan State University
Tags: limit laws, calculus I, MTH 132, direct substitution, factoring limits, rationalising, indeterminate forms, 0/0, difference quotient, MSU
Difficulty: Introductory | Prerequisites: Basic understanding of what a limit is (Sections 1.4–1.5), comfort with factoring polynomials and simplifying rational expressions.
Limit laws are the workhorse rules that let you break a complicated limit into smaller, manageable pieces. Rather than proving every limit from the epsilon-delta definition, these laws give you a toolkit: you can add, subtract, multiply, divide and take roots of limits you already know. This section sits right after you have built intuition about what limits are (Sections 1.4–1.5), and it gives you the mechanical skills you will rely on for the rest of Calculus I.
If you missed the last few weeks, the key prior knowledge is this: a limit describes the value a function approaches as the input approaches some target, and a polynomial's limit can be found by plugging in. Section 1.6 formalises that intuition into a set of rules and then teaches you what to do when plugging in fails (the 0/0 indeterminate form).
Limit laws let you split a limit of a sum, product, quotient or root into the limits of the individual pieces. When direct substitution gives you a real number, that number is the limit. When substitution gives 0/0, you need to do algebra first (factor, rationalise, simplify) to cancel the problem term, then substitute again.
Limit law
A rule that relates the limit of a combined expression to the limits of its component parts, provided those component limits exist. Think of it as permission to "pull the limit inside" an operation.
Direct substitution
The technique of evaluating a limit by plugging the target value straight into the function. In simple terms, if the function is continuous at that point, you just substitute and you are done.
Indeterminate form (0/0)
The result you get when direct substitution gives 0 in both the numerator and denominator. It does not mean the limit is zero or undefined; it means you cannot tell yet and need to do more work. Think of it as a "try again" signal.
Factoring (for limits)
Rewriting a numerator or denominator as a product of simpler expressions so that the common factor causing 0/0 can be cancelled. In simple terms, you break things apart to remove the piece that is causing the problem.
Rationalising (conjugate multiplication)
Multiplying numerator and denominator by the conjugate of a square-root expression to eliminate the radical. This converts a difference of roots into a difference of squares, which can then be simplified.
Difference quotient
An expression of the form [f(a + h) – f(a)] / h. This is the skeleton of the derivative definition, and evaluating its limit as h approaches 0 gives the instantaneous rate of change at a.
If lim(x→a) f(x) = L and lim(x→a) g(x) = M both exist, then:
Sum law: lim(x→a) [f(x) + g(x)] = L + M
Difference law: lim(x→a) [f(x) – g(x)] = L – M
Product law: lim(x→a) [f(x) · g(x)] = L · M
Constant multiple law: lim(x→a) [c · f(x)] = c · L
Quotient law: lim(x→a) [f(x) / g(x)] = L / M, provided M ≠ 0
Power law: lim(x→a) [f(x)]^n = L^n
Root law: lim(x→a) ⁿ√f(x) = ⁿ√L (when the root is defined)
What happens when M = 0 in the quotient law?
The quotient law does not apply. If L ≠ 0 and M = 0, the limit does not exist (there is a vertical asymptote). If both L = 0 and M = 0, you have an indeterminate form and must use algebra to simplify before trying again.
Worked example (matching Problem 1):
Given lim(x→a) f(x) = 1, lim(x→a) g(x) = –3, lim(x→a) h(x) = 0:
lim [f + g] = 1 + (–3) = –2 (sum law)
lim [f – g] = 1 – (–3) = 4 (difference law)
lim [f · g] = 1 · (–3) = –3 (product law)
lim [f / g] = 1 / (–3) = –1/3 (quotient law, denominator ≠ 0)
lim [h / g] = 0 / (–3) = 0 (quotient law, denominator ≠ 0)
lim [g / h] = –3 / 0 → does not exist (quotient law fails, numerator ≠ 0)
lim √f = √1 = 1 (root law)
lim g⁻¹ = 1/(–3) = –1/3 (quotient law applied to 1/g)
lim 1/(f + h) = 1/(1 + 0) = 1 (sum law, then quotient law)
For any polynomial p(x), lim(x→a) p(x) = p(a). Just plug in.
For any rational function p(x)/q(x), lim(x→a) [p(x)/q(x)] = p(a)/q(a), provided q(a) ≠ 0.
Worked example (Problem 2):
lim(x→–3) (x – 7) / (7x² – 7x + 5)
Plug in x = –3:
Numerator: –3 – 7 = –10
Denominator: 7(9) – 7(–3) + 5 = 63 + 21 + 5 = 89
Answer: –10/89
The denominator is not zero, so direct substitution works.
Worked example (Problem 3):
lim(x→6) (6x² + 5)(6x + 6)
This is a product of two polynomials. Substitute x = 6:
First factor: 6(36) + 5 = 216 + 5 = 221
Second factor: 6(6) + 6 = 36 + 6 = 42
Answer: 221 × 42 = 9282
When direct substitution gives 0/0, the limit may still exist. Your job is to simplify algebraically until the 0/0 goes away, then substitute.
Technique 1: Factor and cancel
Used when both numerator and denominator are polynomials that share a common factor.
Worked example (Problem 4):
lim(x→3) (x² – 9) / (x – 3)
Substituting gives 0/0. Factor the numerator: x² – 9 = (x – 3)(x + 3).
Cancel (x – 3): lim(x→3) (x + 3) = 3 + 3 = 6
Worked example (Problem 5):
lim(x→1) (x – 1) / (x² + 5x – 6)
Substituting gives 0/0. Factor the denominator: x² + 5x – 6 = (x – 1)(x + 6).
Cancel (x – 1): lim(x→1) 1/(x + 6) = 1/(1 + 6) = 1/7
Worked example (Problem 6):
lim(x→2) (x² + 7x – 18) / (x² – 2x)
Substituting gives 0/0. Factor both:
Numerator: (x + 9)(x – 2)
Denominator: x(x – 2)
Cancel (x – 2): lim(x→2) (x + 9)/x = (2 + 9)/2 = 11/2
Technique 2: Rationalise (conjugate multiplication)
Used when a square root appears and direct substitution gives 0/0. Multiply top and bottom by the conjugate.
Worked example (Problem 7):
lim(t→25) (√t – 5) / (t – 25)
Substituting gives 0/0. Notice t – 25 = (√t – 5)(√t + 5).
Cancel (√t – 5): lim(t→25) 1/(√t + 5) = 1/(5 + 5) = 1/10
Worked example (Problem 10):
lim(a→81) (81 – a) / (9 – √a)
Substituting gives 0/0. Write 81 – a = (9 – √a)(9 + √a).
Cancel (9 – √a): lim(a→81) (9 + √a) = 9 + 9 = 18
Technique 3: Simplify a complex fraction
Used when the expression has fractions within a fraction.
Worked example (Problem 9):
lim(h→0) [7/(–3 + h) + 7/3] / h
Substituting gives 0/0. Combine the fractions in the numerator over a common denominator 3(–3 + h):
7·3 + 7·(–3 + h) all over 3(–3 + h) = [21 – 21 + 7h] / [3(–3 + h)] = 7h / [3(–3 + h)]
Divide by h: 7 / [3(–3 + h)].
Now substitute h = 0: 7 / [3(–3)] = 7 / (–9) = –7/9
These limits look like [f(a + h) – f(a)] / h as h → 0. They preview the derivative, which you will formalise in Chapter 2. The strategy is to expand, simplify, cancel h, then substitute h = 0.
Worked example (Problem 8):
lim(h→0) [2(4 + h)² – 32] / h
Expand: 2(16 + 8h + h²) – 32 = 32 + 16h + 2h² – 32 = 16h + 2h².
Factor: h(16 + 2h).
Cancel h: lim(h→0) (16 + 2h) = 16 + 0 = 16
Note: this is the derivative of 2x² evaluated at x = 4. You will see this connection made explicit soon.
The core limit laws (given lim f(x) = L and lim g(x) = M):
Sum: lim [f ± g] = L ± M
Product: lim [f · g] = L · M
Quotient: lim [f / g] = L / M (M ≠ 0)
Power: lim [f]^n = L^n
Root: lim ⁿ√f = ⁿ√L (when defined)
Constant: lim c = c
Identity: lim x = a
Direct substitution property:
For any polynomial or rational function (with non-zero denominator at the point), lim(x→a) f(x) = f(a).
Key algebraic identities used in this section:
Difference of squares: a² – b² = (a – b)(a + b)
This works with roots: t – 25 = (√t – 5)(√t + 5)
Limit laws are the mathematical basis for how engineers and scientists model smooth change. Whenever you calculate an instantaneous velocity, a reaction rate, or a marginal cost, you are evaluating a limit of a difference quotient, and the algebraic techniques from this section are exactly what make that calculation possible.
Students often think 0/0 means the limit is 0 or does not exist. It does not mean either. It is an indeterminate form that requires further work.
Students sometimes apply the quotient law when the denominator's limit is 0. The quotient law explicitly requires M ≠ 0. When M = 0, you must simplify first.
When rationalising, students sometimes multiply only the numerator by the conjugate and forget the denominator. You must multiply both top and bottom to keep the expression equal to the original.
Students occasionally cancel terms that are added rather than multiplied. You can only cancel a common factor, not a common term. (x + 3) cancels from (x + 3)(x – 1) / (x + 3), but the 3 in (x + 3)/(x + 3) is not being added in the right way to cancel one piece of the sum.
⚠️ Expect at least one problem where direct substitution gives 0/0 and you must factor or rationalise before evaluating. This is the most common limit question on early exams.
⚠️ Know when the quotient law fails. If substitution gives a non-zero numerator over zero, the limit does not exist (vertical asymptote). If it gives 0/0, you need algebra.
⚠️ Difference-quotient limits (the [f(a+h) – f(a)]/h form) preview the derivative definition. Expect these to reappear in Chapter 2, so build fluency now.
True or False: If lim(x→a) g(x) = 0, then lim(x→a) f(x)/g(x) does not exist.
False. It depends on whether lim(x→a) f(x) is also 0. If so, it is indeterminate and might still have a finite limit.
Fill in the blank: lim(x→5) (x² – 25)/(x – 5) = ___
Factor: (x – 5)(x + 5)/(x – 5) = x + 5. Substitute: 10.
True or False: You can always evaluate a limit by plugging in.
False. Direct substitution works only when the function is defined and continuous at the point. When it produces 0/0, algebraic simplification is needed first.
Fill in the blank: lim(h→0) [(3 + h)² – 9] / h = ___
Expand: 9 + 6h + h² – 9 = 6h + h² = h(6 + h). Cancel h. Substitute: 6.
Q: Given lim(x→a) f(x) = 4 and lim(x→a) g(x) = –2, find lim(x→a) [3f(x) – g(x)²].
A: Apply the limit laws: 3(4) – (–2)² = 12 – 4 = 8.
Q: Evaluate lim(x→–1) (x² – 1)/(x + 1).
A: Substitution gives 0/0. Factor: (x – 1)(x + 1)/(x + 1). Cancel (x + 1). lim(x→–1) (x – 1) = –1 – 1 = –2.
Q: Evaluate lim(x→4) (√x – 2)/(x – 4).
A: Substitution gives 0/0. Write x – 4 = (√x – 2)(√x + 2). Cancel (√x – 2). lim(x→4) 1/(√x + 2) = 1/(2 + 2) = 1/4.
Q: Evaluate lim(h→0) [5(1 + h)³ – 5] / h.
A: Expand: 5(1 + 3h + 3h² + h³) – 5 = 15h + 15h² + 5h³ = h(15 + 15h + 5h²). Cancel h. Substitute h = 0: 15.
Q: If lim(x→a) f(x) = 2 and lim(x→a) g(x) = 0, does lim(x→a) f(x)/g(x) exist?
A: No. The numerator approaches 2 (non-zero) while the denominator approaches 0, so the quotient law does not apply and the limit does not exist.
This connects directly to continuity (Section 1.8): a function is continuous at a point precisely when you can evaluate its limit by direct substitution. The algebraic techniques here (factoring, rationalising) are the same ones you will use to find derivatives from the limit definition in Chapter 2. The difference-quotient problems in this assignment are, in fact, derivative calculations in disguise.
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