Difficulty: Introductory to Intermediate | Prerequisites: Section 4–5 study notes (limits, one-sided limits, vertical asymptotes).
Section 6 gives you the algebraic machinery for computing limits by hand. Most limit problems fall into three cases based on what happens when you try direct substitution: you get a real number (done), you get N/0 (vertical asymptote), or you get the indeterminate form 0/0 (apply algebra, then retry). The Squeeze Theorem handles the tricky cases that algebra alone cannot crack. Section 8 introduces continuity, the property that lets you skip the limit process and simply plug in, then classifies the ways a function can fail to be continuous, and closes with the Intermediate Value Theorem.
When you plug the target value into a function and get a normal number, that number is the limit (direct substitution). When you get 0/0 instead, use algebra (factor, multiply by a conjugate, find a common denominator) to simplify, then substitute again. Continuity is the formal name for "direct substitution works," and the Intermediate Value Theorem guarantees that a continuous function hits every y-value between its endpoints.
Limit laws
A set of rules that let you break a complicated limit into simpler pieces. For example, the limit of a sum equals the sum of the limits, the limit of a product equals the product of the limits, and so on, provided each individual limit exists.
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). In simple terms, for these well-behaved functions, "just plug it in."
Indeterminate form (0/0)
When direct substitution yields 0 in both the numerator and denominator. This does not mean the limit is 0 or undefined; it means more work is needed. Factor, multiply by a conjugate, or find a common denominator, then try again.
Think of it as a signal that the numerator and denominator are both heading to zero at the same rate, and you need algebra to untangle what happens.
Squeeze Theorem (Sandwich 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.
In simple terms, if a function is trapped between two others that both converge to the same value, the trapped function must converge there too.
Continuous function
A function f is continuous at x = a if three conditions hold: f(a) is defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a). Informally, you can draw the graph through x = a without lifting your pencil.
Removable discontinuity
A discontinuity that can be "fixed" by redefining the function at a single point. Graphically, it looks like a hole in the curve. Typically arises from the 0/0 case after cancellation.
Infinite discontinuity
Another name for a vertical asymptote. The function blows up to ±∞ near that point. Typically arises from the N/0 case.
Jump discontinuity
A discontinuity where the left-hand and right-hand limits both exist but are not equal. The graph "jumps" from one value to another. Common in piecewise-defined functions.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any number between f(a) and f(b), then there exists at least one c in (a, b) where f(c) = N.
In simple terms, a continuous function cannot skip over a value. If f starts below a target and ends above it (or vice versa), it must hit that target somewhere in between.
When you first try direct substitution on lim(x→a) f(x)/g(x), the result sorts into one of three cases:
Case 1 (0/N or N/N, where N ≠ 0): Direct substitution gives a finite number. You are done; that number is the limit. This works because of the Direct Substitution Property for polynomials, rationals, and root functions in their domains.
Case 2 (N/0): The numerator approaches a nonzero number while the denominator approaches zero. This signals a vertical asymptote. Determine whether the limit is +∞ or -∞ by checking the sign of both numerator and denominator from each side.
Case 3 (0/0): Both numerator and denominator approach zero. This is an indeterminate form. You must simplify algebraically before substituting again.
Factoring and cancelling: Factor both numerator and denominator, cancel the common factor that causes the 0/0, then substitute. 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.
Multiplying by a conjugate: For limits involving square roots, multiply numerator and denominator by the conjugate. 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) → 1/6.
Common denominators: When the expression involves compound fractions, combine using a common denominator, simplify, then substitute.
Cancellation theorem: If f(x) = g(x) for all x ≠ a, then lim(x→a) f(x) = lim(x→a) g(x), provided the limit exists. This justifies why cancelling a factor that equals zero at x = a does not change the limit.
Used when algebraic simplification is not enough (for example, limits involving sin(x)/x).
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.
Classic application: 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 the limit is 0.
Another classic: if 2x + 1 ≤ g(x) ≤ x² - 2x + 5 for all x, and you need lim(x→2) g(x), evaluate the bounds at x = 2: both give 5, so by Squeeze, the limit is 5.
A function f is continuous at x = a when: (i) f(a) is defined, (ii) lim(x→a) f(x) exists, (iii) lim(x→a) f(x) = f(a).
Polynomials, rational functions (in their domain), root functions (in their domain), and trigonometric functions (in their domain) are all continuous everywhere in their domains.
Combinations of continuous functions (sums, differences, products, quotients where the denominator is nonzero, and compositions) are also continuous. This means lim(x→a) f(g(x)) = f(lim(x→a) g(x)) when both functions are continuous at the relevant points.
Removable discontinuity (hole): The limit exists but either f(a) is not defined or f(a) ≠ the limit. Typically from the 0/0 case. You can "remove" it by redefining f(a) to equal the limit.
Infinite discontinuity (vertical asymptote): At least one one-sided limit is ±∞. Typically from the N/0 case.
Jump discontinuity: Both one-sided limits exist and are finite, but they are not equal. Typically from piecewise-defined functions.
Watch for: dividing by zero, square roots of negatives, and boundaries of piecewise definitions.
Statement: if f is continuous on [a, b] and N is any number strictly between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N.
The IVT does not tell you what c is. It guarantees existence only.
Common exam use: show a root exists. Evaluate f at two points; if f changes sign (one positive, one negative), IVT guarantees a root between them. Example: show x⁴ + x - 3 = 0 has a solution in (1, 2). f(1) = -1 < 0, f(2) = 15 > 0, and f is continuous (polynomial), so by IVT some c in (1, 2) satisfies f(c) = 0.
Another pattern: prove cos(x) = x³ has a solution. Let f(x) = cos(x) - x³. f(0) = 1 > 0, f(π/2) ≈ -3.9 < 0. Since f is continuous, IVT guarantees a c in (0, π/2) where f(c) = 0.
Limit laws (assuming lim(x→a) f(x) and lim(x→a) g(x) both exist, and c is a constant):
lim [f(x) + g(x)] = lim f(x) + lim g(x)
lim [c · f(x)] = c · lim f(x)
lim [f(x) · g(x)] = lim f(x) · lim g(x)
lim [f(x) / g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0
lim c = c
lim x = a
lim xⁿ = aⁿ
lim [f(x)]ⁿ = [lim f(x)]ⁿ
\text{Continuity at } a \colon \quad \lim_{x \to a} f(x) = f(a)\textbf{Squeeze Theorem:} \quad f(x) \leq g(x) \leq h(x) \text{ near } a,\; \lim_{x\to a} f(x) = \lim_{x\to a} h(x) = L \;\Longrightarrow\; \lim_{x\to a} g(x) = L\textbf{IVT:} \quad f \text{ continuous on } [a,b],\; N \text{ between } f(a) \text{ and } f(b) \;\Longrightarrow\; \exists\, c \in (a,b) \text{ with } f(c) = NContinuity underlies every physical measurement that varies smoothly: temperature, altitude, pressure. The IVT is used in engineering root-finding algorithms (bisection method) to narrow down where a function crosses zero. In economics, the IVT guarantees that a continuously changing price must pass through every intermediate value.
Students often assume 0/0 means the limit is 0 or does not exist. It is an indeterminate form, meaning the answer could be any number (or ±∞). You must simplify first.
Assuming that continuity at every point in a domain means "no breaks" on the number line. Rational functions like 1/(x - 2) are continuous on their domain, which excludes x = 2. The function is continuous everywhere it is defined; it simply is not defined at x = 2.
Confusing "the limit does not exist" with "I cannot compute it." A limit genuinely does not exist when the left-hand and right-hand limits disagree, or when the function oscillates without settling.
Applying the IVT without first confirming continuity on the closed interval. The theorem fails for discontinuous functions.
⚠️ "Plug in first" is always the first step. The result tells you which case you are in. Students who jump straight to factoring without checking often waste time or misidentify the problem type.
⚠️ The 0/0 algebraic techniques (factor, conjugate, common denominator) are heavily tested. Know all three and recognise which to apply by the shape of the expression.
⚠️ Squeeze Theorem problems are a common exam question. The key step is bounding the tricky part (usually a trig function) between -1 and 1, then multiplying through by the vanishing factor.
⚠️ For continuity of piecewise functions, check each boundary point: evaluate both one-sided limits and compare to the function value at the boundary.
⚠️ IVT proofs follow a fixed template: state f is continuous, evaluate f(a) and f(b), note the sign change, conclude by IVT.
True or False: If direct substitution gives 0/0, the limit does not exist.
Fill in the blank: To evaluate lim(x→4) (x - 4)/(√x - 2), multiply numerator and denominator by ___.
True or False: Every polynomial is continuous on (-∞, ∞).
True or False: A removable discontinuity always involves a vertical asymptote.
Fill in the blank: To apply the IVT to show a root exists, you need f to be ___ on a closed interval, with f(a) and f(b) having opposite signs.
Answers: 1. False (it is indeterminate; more work is needed). 2. (√x + 2), the conjugate. 3. True. 4. False (a removable discontinuity is a hole, not an asymptote). 5. Continuous.
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)] = (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: 1/6.
Q: Use the Squeeze Theorem to find lim(x→0) x² cos(2/x).
A: Since -1 ≤ cos(2/x) ≤ 1, we have -x² ≤ x² cos(2/x) ≤ x². Both bounds approach 0, so by the Squeeze Theorem the limit is 0.
Q: Determine and classify the discontinuities of f(x) = (x² + 5x + 6)/(x² - 3x - 10).
A: Factor: (x + 2)(x + 3)/[(x - 5)(x + 2)]. Discontinuities at x = 5 and x = -2. At x = -2, the (x + 2) cancels, giving a removable discontinuity (hole). At x = 5, the numerator is nonzero (8 · 8/0), giving an infinite discontinuity (vertical asymptote).
Q: Use the IVT to show that x⁴ + x - 3 = 0 has a root in (1, 2).
A: Let f(x) = x⁴ + x - 3. f is a polynomial, hence continuous everywhere. f(1) = 1 + 1 - 3 = -1 < 0. f(2) = 16 + 2 - 3 = 15 > 0. Since f(1) < 0 < f(2), by the IVT there exists c in (1, 2) with f(c) = 0.
The limit laws and direct substitution property carry directly into computing derivatives in Chapter 2, since the derivative is itself a limit. Continuity is a prerequisite for differentiability (if a function is differentiable at a point, it must be continuous there, but not vice versa). The IVT comes back in later courses as the foundation for the bisection method and for proving existence theorems in analysis.
limit laws, direct substitution, indeterminate form, 0/0, factoring limits, conjugate limits, squeeze theorem, sandwich theorem, pinching theorem, continuous function, continuity, removable discontinuity, hole, infinite discontinuity, vertical asymptote, jump discontinuity, piecewise continuity, Intermediate Value Theorem, IVT, bisection, root finding, MTH 132, Calculus I, Chapter 1, Section 1.6, Section 1.8, Stewart Calculus