Source: MTH 132, Chapter 1 (Sections 6.1–6.5) | Michigan State University
Tags: limit laws, direct substitution, factoring limits, conjugate, Squeeze Theorem, 0/0 indeterminate form, cancellation, calculus 1, MTH 132, MSU
Difficulty: Intermediate. Prerequisites: intuitive definition of a limit (Section 5), basic algebra (factoring, rationalising, absolute values).
Section 5 taught you what a limit is. This section teaches you how to compute one without a graph or a table. You will learn a systematic approach: try direct substitution first, then classify the result. If you get a number over a number, you are done. If you get 0/nonzero (or nonzero/0), those are specific cases. If you get 0/0, that is the indeterminate form and you need algebraic manipulation (factoring, conjugates, common denominators) before trying substitution again. The Squeeze Theorem handles the cases algebra cannot reach.
Always start by plugging in. If the result is a clean fraction (no zero denominator), that is the limit. If you land on 0/0, simplify algebraically and try again. If the denominator is zero but the numerator is not, you are looking at a vertical asymptote (±∞ or DNE). The Squeeze Theorem pins down limits of oscillating or bounded functions by trapping them between two functions whose limits you already know.
Limit laws
A set of rules that let you break a complicated limit into simpler pieces: sum, difference, constant multiple, product, quotient, power, and root laws. They apply whenever the individual limits exist.
In simple terms, you can "distribute" the limit across addition, multiplication, and so on, as long as each piece has a limit.
Direct substitution property
If f is a polynomial, rational function, or root function and a is in the domain of f, then lim(x→a) f(x) = f(a). Just plug in.
In simple terms, for "nice" functions, the limit is just the function value.
Indeterminate form (0/0)
When direct substitution gives 0/0, the limit could be any number (or ±∞ or DNE). You cannot determine the answer from 0/0 alone; further work is required.
In simple terms, 0/0 is a sign that says "keep going, you are not done yet."
Cancellation theorem
If f(x) = g(x) for all x ≠ a, then lim(x→a) f(x) = lim(x→a) g(x), provided the latter limit exists. This justifies cancelling a common factor that is zero at a.
Squeeze Theorem (Sandwich Theorem, Pinching Theorem)
If f(x) ≤ g(x) ≤ h(x) for all x near a (except possibly at a), and lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L.
In simple terms, if a function is trapped between two others that both approach the same value, it has no choice but to approach that value too.
Always try direct substitution first. Plug a into the function.
If the denominator is nonzero, the limit equals the resulting fraction. Done.
This works for all polynomials, rational functions at points in their domain, and root functions at points in their domain.
Example: lim(x→0) (2x² – x)/(3 + x) = (0 – 0)/(3 + 0) = 0/3 = 0.
Example: lim(x→–3) √(1 – x)/(x² – 4x) = √4/(9 + 12) = 2/21.
When substitution gives a nonzero numerator and a zero denominator, the limit is ±∞ or does not exist.
Check the sign of the numerator and the sign of the denominator from each side to determine +∞ or –∞.
Example: lim(x→1⁺) –4/(1 – x). Numerator is –4 (negative). As x → 1⁺, (1 – x) is small and negative. Negative divided by small negative = large positive, so the limit is +∞.
The 0/0 form is indeterminate. You must simplify before substituting again.
Factoring: factor numerator and denominator, cancel the common factor. Example: lim(x→1) (x² + 2x – 3)/(x² – 1) = lim(x→1) (x + 3)(x – 1)/[(x + 1)(x – 1)] = lim(x→1) (x + 3)/(x + 1) = 4/2 = 2.
Common denominators: for compound fractions. Example: lim(x→2) [(1/(x–1)) – 1]/(x – 2). Combine the numerator into a single fraction first, then simplify.
Conjugate multiplication: for expressions involving square roots. Example: lim(h→0) (√(9+h) – 3)/h. Multiply top and bottom by (√(9+h) + 3) to get h/[h(√(9+h) + 3)] = 1/(√(9+h) + 3), then substitute h = 0 to get 1/6.
Absolute value limits: rewrite |x – a| as (x – a) or –(x – a) depending on the direction of approach. Example: lim(t→1⁺) |t – 1|/(1 – t) = (t – 1)/(1 – t) = –1.
Used when algebraic simplification is not possible, typically with oscillating functions like sin(1/x) or cos(2/x) multiplied by a vanishing factor.
Strategy: find a lower bound f(x) and an upper bound h(x) for g(x), both with the same limit at a.
Example: lim(x→0) x² cos(2/x). Since –1 ≤ cos(2/x) ≤ 1, we have –x² ≤ x² cos(2/x) ≤ x². Both –x² and x² approach 0 as x → 0, so by the Squeeze Theorem, the limit is 0.
Example: if 2x + 1 ≤ g(x) ≤ x² – 2x + 5 for all x, find lim(x→2) g(x). Both bounds give 5 at x = 2, so g(x) → 5.
The eight limit laws (assuming lim(x→a) f(x) and lim(x→a) g(x) both exist and c is a constant):
Sum: lim(x→a) [f(x) + g(x)] = lim f(x) + lim g(x)
Constant multiple: lim(x→a) [c · f(x)] = c · lim f(x)
Product: lim(x→a) [f(x) · g(x)] = lim f(x) · lim g(x)
Quotient: lim(x→a) [f(x)/g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0
Constant: lim(x→a) c = c
Identity: lim(x→a) x = a
Power: lim(x→a) xⁿ = aⁿ (when appropriate)
Power of a function: lim(x→a) [f(x)]ⁿ = [lim f(x)]ⁿ (when appropriate)
Direct substitution property: if f is a polynomial, rational, or root function and a is in the domain, then lim(x→a) f(x) = f(a).
Squeeze Theorem: if f(x) ≤ g(x) ≤ h(x) near a and lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L.
The Squeeze Theorem is used in signal processing to show that noisy oscillating signals decay to zero when multiplied by a shrinking envelope. In engineering, limit computations underpin error analysis: showing that a measurement error vanishes as the sample size grows. The 0/0 algebraic technique of factoring and cancelling is the same logic used in simplifying transfer functions in control theory.
Students often try to "evaluate" 0/0 as if it equals 0 or 1 or ∞. It is none of those. 0/0 is indeterminate, meaning you need more work.
A frequent error is cancelling a factor without checking that it is zero at the limit point. You can only cancel a common factor of (x – a) when the cancellation theorem applies (i.e. f(x) = g(x) for x ≠ a).
Students sometimes apply the quotient law when the denominator's limit is zero. The quotient law explicitly requires lim g(x) ≠ 0.
When using the Squeeze Theorem, students occasionally bound the function incorrectly (e.g. writing –x ≤ x² cos(2/x) instead of –x² ≤ x² cos(2/x)). The bounds must be valid for all x near a.
⚠️ The three-case classification (N/N, N/0, 0/0) is the single most important exam strategy in this section. Memorise the decision process.
⚠️ Factoring limits (the 0/0 case) appear on virtually every MTH 132 exam. Practise factoring quadratics quickly.
⚠️ Conjugate multiplication problems (limits involving square roots) are a favourite exam question. Know the technique cold.
⚠️ The Squeeze Theorem is typically tested with x² times a bounded trigonometric function. Expect at least one such problem.
Fill in the blank: the first step in evaluating any limit is to try ____.
Direct substitution.
True or false: 0/0 means the limit is 0.
False. 0/0 is indeterminate.
Fill in the blank: if direct substitution gives N/0 (nonzero over zero), the function likely has a ____ at that point.
Vertical asymptote.
True or false: the Squeeze Theorem can be used when f(x) ≤ g(x) ≤ h(x) and lim f(x) = lim h(x).
True.
What algebraic technique handles lim(h→0) (√(9+h) – 3)/h?
Multiply by the conjugate (√(9+h) + 3) / (√(9+h) + 3).
Q: Evaluate lim(x→1) (x² + 2x – 3)/(x² – 1).
A: Direct substitution gives 0/0. Factor: (x + 3)(x – 1) / [(x + 1)(x – 1)]. Cancel (x – 1): (x + 3)/(x + 1). Substitute x = 1: 4/2 = 2.
Q: Evaluate lim(h→0) (√(9 + h) – 3)/h.
A: Direct substitution gives 0/0. Multiply by conjugate: h / [h(√(9+h) + 3)] = 1/(√(9+h) + 3). Substitute h = 0: 1/(3 + 3) = 1/6.
Q: Evaluate lim(x→0) x² cos(2/x).
A: cos(2/x) oscillates but is bounded: –1 ≤ cos(2/x) ≤ 1. So –x² ≤ x² cos(2/x) ≤ x². Both bounds → 0 as x → 0. By the Squeeze Theorem, the limit is 0.
Q: If lim(x→2) f(x) = 3 and lim(x→2) g(x) = –2, find lim(x→2) [5f(x) – x·g(x)].
A: By limit laws: 5(3) – 2(–2) = 15 + 4 = 19.
Q: Evaluate lim(x→9) (√x – 3)/(9 – x).
A: Direct substitution gives 0/0. Factor denominator: 9 – x = –(x – 9) = –(√x – 3)(√x + 3). Cancel (√x – 3): 1/[–(√x + 3)]. Substitute x = 9: 1/(–6) = –1/6.
The limit laws and algebraic techniques here feed directly into Section 8 on continuity (a function is continuous when direct substitution works) and into the definition of the derivative in Chapter 2, which is itself a limit of the form lim(h→0) [f(a+h) – f(a)]/h, a 0/0 indeterminate form that requires exactly the algebraic skills practised here. The Squeeze Theorem returns when you prove lim(x→0) sin(x)/x = 1 in later sections.
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limit laws, direct substitution, direct substitution property, indeterminate form, 0/0, factoring limits, conjugate multiplication, rationalising, common denominator limit, cancellation theorem, Squeeze Theorem, Sandwich Theorem, Pinching Theorem, polynomial limit, rational function limit, MTH 132, Calculus 1, MSU, Michigan State, Chapter 1