Difficulty: Foundational | Prerequisites: Algebra, rational expressions, factoring
Limits and continuity form the backbone of everything in calculus. Before you can differentiate or integrate, you need to know how a function behaves as x approaches a given value, and whether the function is "well-behaved" (continuous) at that point. This unit sits right at the start of MAC2233 and every topic that follows, from derivatives to optimisation, depends on it. If you are coming in cold, make sure your algebra (especially factoring and rational expressions) is solid.
A limit describes the value a function approaches as x gets close to some number, even if the function is not defined there. Continuity means the function has no holes, jumps, or vertical asymptotes at a point. Most limit problems in this course reduce to algebra: factor, cancel, then plug in.
Limit
The value that f(x) approaches as x gets arbitrarily close to a particular number c. Written lim(x→c) f(x). Think of it as: "What y-value is the function heading toward, even if it never lands there?"
One-sided limit
The value f(x) approaches from only one direction. lim(x→c⁻) means approaching from the left; lim(x→c⁺) means from the right. In simple terms, this means you only care about what happens on one side of c.
Limit at infinity
The behaviour of f(x) as x grows without bound (x→∞) or decreases without bound (x→-∞). You find it by comparing the highest-power terms in the numerator and denominator.
Continuity
A function f(x) is continuous at x = c when three conditions hold: f(c) exists, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c). Think of it as: "You can draw the graph through that point without lifting your pen."
Discontinuity
A point where at least one of the three continuity conditions fails. Common types: holes (removable), jumps, and vertical asymptotes.
Indeterminate form (0/0)
When direct substitution gives 0/0, the limit is not automatically zero or undefined. It means more work is needed (factor, rationalise, simplify) before the true value emerges.
Average rate of change
The slope of the secant line between two points on a curve: [f(x₁) - f(x₂)] / [x₁ - x₂]. In simple terms, this means how fast the output changes per unit of input over an interval.
The first thing to try: plug the target value into f(x).
If you get an actual number, that number is the limit. Done.
If you get 0/0, the limit is indeterminate and you need another technique.
When direct substitution gives 0/0, factor both the numerator and denominator.
Cancel the common factor that is producing the zero.
Then substitute again.
Example from the course: lim(x→-3) of (x² - 2x - 15)/(x² + 2x - 3). Factor to get (x-5)(x+3) / (x+3)(x-1). Cancel (x+3), leaving (x-5)/(x-1). Plug in x = -3: (-3-5)/(-3-1) = -8/-4 = 2.
When the expression contains a square root and direct substitution gives 0/0, multiply numerator and denominator by the conjugate.
Example from the course: lim(x→4) of (√x - 2)/(x - 4). Multiply top and bottom by (√x + 2). The numerator becomes x - 4, which cancels with the denominator, leaving 1/(√x + 2). Plug in x = 4: 1/(2+2) = 1/4.
Compare the degree of the numerator (n) and the degree of the denominator (d).
If n < d, the limit is 0.
If n = d, the limit is the ratio of the leading coefficients.
If n > d, the limit is ±∞ (determine the sign from the leading terms).
Example: lim(x→∞) of (-2x+1)/(1+2x+2x²). Leading terms: -2x / 2x² = -1/x → 0.
Example: lim(x→-∞) of (1-3x+4x²)/(5-2x²). Leading terms: 4x²/(-2x²) = -2.
If the denominator approaches 0 but the numerator does not, the limit is ±∞.
Determine the sign by checking whether the numerator and denominator are both positive, both negative, or mixed as x approaches the target.
Example: lim(x→-3⁻) of (6-x)/(x+3). As x→-3 from the left, the numerator → 9 (positive) and the denominator → 0⁻ (negative). Positive / negative = -∞.
Formula: [f(x₁) - f(x₂)] / (x₁ - x₂)
Between x and x+h: [f(x+h) - f(x)] / h
Example: for f(x) = 5 - x - 2x², the average rate of change between x and x+h simplifies to -1 - 4x - 2h.
For a numerical interval (say x = 1 to x = 0.8), compute f at each endpoint and divide by the change in x.
Polynomials are continuous everywhere (all real numbers, ℝ).
Rational functions (fractions of polynomials) are continuous everywhere except where the denominator equals zero.
To find discontinuities of a rational function, set the denominator equal to zero and solve.
f(x) = 4x/(4x - x²) = 4x/[x(4-x)]. Denominator is zero when x = 0 or x = 4. Continuous on ℝ except x = 0 and x = 4.
f(x) = (x-2)/(x²+1). The denominator x²+1 is always positive (no real roots), so this function is continuous on all of ℝ.
f(x) = 4x/(x²-4x) = 4/[x(x-4)] = 4/(x-4) after cancelling x. Continuous on ℝ except x ≠ 4 (and technically x ≠ 0 from the original form).
f(x) = (x-3)/(x²+9). The denominator x²+9 is always positive, so continuous on all of ℝ.
For a piecewise function to be continuous at the boundary point, the left-hand limit and the right-hand limit must both equal the function value at that point.
Example: f(x) = { -2x² + ax + 1 for x < 1; 3x + 1 for x ≥ 1 }. For continuity at x = 1, set lim(x→1⁻) = lim(x→1⁺). Left side: -2(1)² + a(1) + 1 = a - 1. Right side: 3(1) + 1 = 4. So a - 1 = 4, giving a = 5.
Average rate of change: [f(x₁) - f(x₂)] / (x₁ - x₂)
Difference quotient: [f(x+h) - f(x)] / h
Limits at infinity for rational functions: compare highest-degree terms in numerator and denominator
Continuity test (three conditions): (1) f(c) is defined, (2) lim(x→c) f(x) exists, (3) lim(x→c) f(x) = f(c)
Piecewise continuity: set left-hand limit = right-hand limit at the boundary and solve for the unknown parameter
Students often think 0/0 means the limit is zero or does not exist. It does not. 0/0 is an indeterminate form, which means the true limit could be any finite number, ∞, or -∞. You have to simplify further.
Students often confuse "the function is undefined at x = c" with "the limit at x = c does not exist." These are different things. A limit can exist even when the function is not defined there (that is the whole point of limits).
When evaluating limits at infinity, students sometimes look at every term instead of just the leading terms. Only the highest-degree terms in the numerator and denominator matter.
In piecewise continuity problems, students sometimes forget that the function value at the boundary must also equal the limit, not just that the two pieces match from either side.
⚠️ Factoring and cancelling is the most commonly tested limit technique in MAC2233. Expect at least two or three questions that require it.
⚠️ Limits at infinity appear on every test. Know the three cases (n < d, n = d, n > d) cold.
⚠️ Piecewise continuity problems ("find the value of a") are a near-guaranteed exam question. The method is mechanical: set left limit = right limit and solve.
⚠️ Average rate of change problems show up both as algebraic (simplify [f(x+h) - f(x)]/h) and as numerical (compute at two given x-values). Be comfortable with both.
⚠️ Determining where a rational function is continuous is a quick-marks question. Set the denominator to zero, list the excluded values, state ℝ except those.
True or false: If f(3) is undefined, then lim(x→3) f(x) does not exist.
False. The limit can exist even when the function is not defined at that point.
Fill in the blank: When direct substitution gives 0/0, the form is called ______.
Indeterminate.
True or false: For the function f(x) = (x²-9)/(x-3), the limit as x→3 equals 6 even though f(3) is undefined.
True. Factor: (x-3)(x+3)/(x-3) = x+3. Plug in 3: 6.
Fill in the blank: If the degree of the numerator is less than the degree of the denominator, the limit as x→∞ is ______.
True or false: A polynomial function is continuous on all real numbers.
True.
Q: Evaluate lim(x→-3) of (x² - 2x - 15)/(x² + 2x - 3).
A: Factor numerator: (x-5)(x+3). Factor denominator: (x+3)(x-1). Cancel (x+3). Plug in x = -3: (-3-5)/(-3-1) = -8/-4 = 2.
Q: Evaluate lim(x→∞) of (3x² - 4x²)/(2x + 6x²).
A: Simplify numerator to -x². Leading terms: -x²/6x² = -1/6. But note the sign: positive divided by positive depends on the leading coefficients. Here the original gives 3x²-4x² = -x², and 6x², so the limit is -1/6.
Q: Determine where f(x) = 4x/(4x - x²) is continuous.
A: Factor denominator: x(4-x). Set to zero: x = 0, x = 4. Continuous on ℝ except x = 0 and x = 4.
Q: Find the value of a that makes f(x) continuous, where f(x) = -2x² + ax + 1 for x < 1 and f(x) = 3x + 1 for x ≥ 1.
A: Left limit at x = 1: -2(1) + a(1) + 1 = a - 1. Right limit: 3(1) + 1 = 4. Set equal: a - 1 = 4, so a = 5.
Q: Find the average rate of change of f(x) = 1 + 2x - 3x² between x and x+h.
A: Compute f(x+h) = 1 + 2(x+h) - 3(x+h)². Expand: 1 + 2x + 2h - 3x² - 6xh - 3h². Subtract f(x): 2h - 6xh - 3h². Divide by h: 2 - 6x - 3h.
Q: Evaluate lim(x→4) of (√x - 2)/(x - 4).
A: Multiply by conjugate (√x + 2)/(√x + 2). Numerator becomes x - 4, cancels with denominator. Left with 1/(√x + 2). Plug in x = 4: 1/4.
Limits lead directly into the definition of the derivative, which is the limit of the difference quotient as h→0. The average rate of change you practise here becomes the instantaneous rate of change (the derivative) in the next unit. Continuity matters again when you study the Extreme Value Theorem and optimisation, because a function must be continuous on a closed interval for the theorem to guarantee absolute max and min values.
Related Terms / Search Tags
limit evaluation, continuity of rational functions, indeterminate form, 0/0, factoring limits, rationalising limits, conjugate method, limits at infinity, leading coefficient test, one-sided limits, piecewise continuity, average rate of change, difference quotient, secant line slope, MAC2233, calculus for business, FIU