Difficulty: Foundational. No prerequisites beyond basic algebra and an understanding of functions.
Limits describe what value a function approaches as the input gets close to some number (or heads to infinity). Continuity means a function has no breaks, jumps, or holes at a given point. Asymptotes are lines the graph gets infinitely close to but never crosses, and they show up when limits head to infinity or when the function value settles at a fixed number as x grows without bound.
Limit
The value that f(x) approaches as x approaches a particular number c. Written as lim (x->c) f(x) = L. In simple terms, this means: "What is the y-value heading towards as x gets really close to c, even if f(c) itself is undefined?"
Continuity at a point
A function f is continuous at x = c if three conditions hold: (1) f(c) is defined, (2) lim (x->c) f(x) exists, and (3) lim (x->c) f(x) = f(c). Think of it as: you can draw the graph through that point without lifting your pen.
One-sided limit
The value f(x) approaches as x comes from one direction only. lim (x->c-) means approaching from the left; lim (x->c+) means approaching from the right. In simple terms: you are only looking at what happens from one side of the point, not both.
Indeterminate form
An expression like 0/0 that does not tell you the limit directly. It signals that algebraic simplification (factoring, rationalising, etc.) is needed before substituting. Think of it as: 0/0 is not an answer, it is a flag saying "do more work."
Horizontal asymptote (HA)
A horizontal line y = L that the graph approaches as x goes to positive or negative infinity. Found by evaluating lim (x->infinity) f(x). In simple terms: the long-run ceiling or floor the function settles towards.
Vertical asymptote (VA)
A vertical line x = a where the function grows without bound (heads to positive or negative infinity). Occurs where the denominator equals zero and the factor does not cancel. Think of it as: a wall the graph shoots up or down along but never touches.
Hole (removable discontinuity)
A point where a factor cancels from numerator and denominator, leaving the function undefined at that x-value even though the limit exists. The graph has a gap (open circle) at that single point. In simple terms: the algebra says the function "should" have a value there, but a cancelled zero prevents it.
Direct substitution: Plug the target value straight into the function. If you get a real number, that is the limit. For constants, lim (x->c) k = k regardless of c.
Indeterminate form (0/0): Factor the numerator and denominator, cancel the common factor, then substitute again.
Example from review Q22b: lim (x->25) (x - 25) / (sqrt(x) - 5). Rationalise by multiplying top and bottom by (sqrt(x) + 5). The expression simplifies to (sqrt(x) + 5), and substituting gives 10.
Limits at infinity: Compare the degrees of numerator and denominator.
Same degree: limit = ratio of leading coefficients. Example from Q22c: lim (x->infinity) (3x^2 - 7x + 1) / (6x^2 + 8) = 3/6 = 1/2.
Numerator degree higher: limit is positive or negative infinity.
Denominator degree higher: limit is 0.
One-sided limits: Evaluate approaching from left (-) or right (+) separately. If the left and right limits disagree, the two-sided limit does not exist.
Example from Q22e: lim (x -> -4+) (x - 3) / (x + 4). Numerator approaches -7. Denominator approaches 0 from the positive side. Result: negative infinity.
Limit of a constant: lim (x->c) k = k. The function f(x) = -30 has limit -30 no matter what x approaches (Q1).
Three conditions at x = c:
f(c) is defined.
lim (x->c) f(x) exists (left and right limits agree).
lim (x->c) f(x) = f(c).
If any one condition fails, the function is discontinuous at that point.
Example from Q28: f(x) is piecewise (2x - 1 for x < 3, -5 for x = 3, x^2 - 4 for x > 3). At x = 3, f(3) = -5 (defined), lim from left = 5, lim from right = 5, so the limit exists and equals 5. But 5 does not equal -5, so condition 3 fails. Not continuous.
Making a piecewise function continuous (Q5): Set the two pieces equal at the boundary. For f(x) = {2x^2 - 3x + 1 if x < -2; Ax - 7 if x >= -2}, substitute x = -2 into both, set them equal, and solve for A. Here 2(4) + 6 + 1 = 15 and -2A - 7 = 15, giving A = -11.
Polynomials are continuous everywhere. Checking continuity of 7x^3 - 5x^2 + 3x - 5 at x = -1 is just confirming the polynomial is defined there (it always is). Statement I in Q2 is true.
Vertical asymptotes appear where the denominator equals zero and the numerator does not (the factor does not cancel).
Holes appear where a factor cancels from both numerator and denominator. The x-coordinate of the hole is the zero of that cancelled factor.
Horizontal asymptotes come from the limit as x approaches infinity.
For f(x) = x^2 / (x - 1), the degree of the numerator (2) exceeds the denominator (1), so there is no horizontal asymptote. The claim in Q2 statement III that the HA is x = 1 is false. (x = 1 is actually the vertical asymptote.)
Example from Q24: f(x) = (x - 5) / (x^2 - 4x - 5). Factor: (x - 5) / ((x - 5)(x + 1)). The (x - 5) cancels, leaving 1/(x + 1). Hole at x = 5 (with y-coordinate 1/6). Vertical asymptote at x = -1. Horizontal asymptote at y = 0.
A limit does not exist when the left-sided and right-sided limits give different values, or when the function oscillates without settling.
From Q2 statement II: lim (x->2) (x - 2) / ((x - 2)(x + 2)). Cancelling gives 1/(x + 2), and substituting x = 2 yields 1/4. So the limit does exist, meaning the original statement ("does not exist") is false.
Limit of a constant: lim (x->c) k = k
Limit of a polynomial: Substitute directly. lim (x->c) p(x) = p(c)
Rational function limits (same degree): lim (x->infinity) (a_n x^n + ...) / (b_n x^n + ...) = a_n / b_n
Rational function limits (higher degree on top): limit is infinity (or negative infinity)
Rational function limits (higher degree on bottom): limit is 0
Rationalisation technique: Multiply numerator and denominator by the conjugate of the radical expression to clear 0/0 forms involving square roots.
Continuity checklist: (1) f(c) defined, (2) lim exists, (3) they match.
Finding holes: Factor fully, cancel common factors. The zero of the cancelled factor is the x-coordinate of the hole. Substitute into the simplified function for the y-coordinate.
Finding vertical asymptotes: After cancelling any common factors, set the remaining denominator equal to zero.
Finding horizontal asymptotes: Evaluate lim (x->infinity) f(x) using the degree comparison rules above.
Limits underpin every rate-of-change calculation in business. When a company asks "what happens to our average cost as production grows very large?" they are evaluating a limit at infinity, which tells them whether economies of scale eventually level off. Continuity matters when modelling pricing or tax brackets: a discontinuous function means a sudden jump in cost that could catch a decision-maker off guard.
Students often think that if f(c) is undefined, the limit at c cannot exist. That is wrong. The limit describes what f(x) approaches, not what f(c) equals. A hole in the graph means f(c) is undefined, yet the limit is perfectly well-defined.
Students frequently confuse vertical asymptotes with horizontal asymptotes. A vertical asymptote is a value of x (written x = a). A horizontal asymptote is a value of y (written y = L). Mixing up the axis is a common exam error.
When the numerator and denominator share a common factor, some students call the zero of that factor a vertical asymptote. It is a hole, not an asymptote. Cancel first, then check what remains.
A 0/0 result from direct substitution does not mean the limit is 0 or does not exist. It means more algebraic work is needed.
⚠️ Expect at least one question requiring you to determine continuity of a piecewise function by checking all three conditions. Show every condition explicitly.
⚠️ Rationalising limits involving square roots (multiplying by the conjugate) is a favourite exam technique. Practise it until it is automatic.
⚠️ "Which statements are true?" questions test whether you can distinguish between asymptotes and holes, and whether limits exist after cancellation.
⚠️ You will likely see a limit at infinity where degree comparison gives the answer in seconds. Know the three cases cold.
⚠️ One-sided limits and their connection to vertical asymptotes often appear in the free-response section.
True or False: lim (x->5) 42 = 42.
True. The limit of a constant function equals that constant, regardless of x.
True or False: If f(3) is undefined, then lim (x->3) f(x) does not exist.
False. A function can have a limit at a point where it is undefined (e.g. a hole).
Fill in the blank: For a rational function, if the degree of the numerator equals the degree of the denominator, the horizontal asymptote is y = ___.
The ratio of the leading coefficients.
True or False: x = 1 is a horizontal asymptote of f(x) = x^2 / (x - 1).
False. x = 1 is a vertical asymptote. Because the numerator has a higher degree, there is no horizontal asymptote.
Fill in the blank: A function is continuous at x = c if f(c) is defined, the limit exists, and ___.
The limit equals f(c).
Q: Evaluate lim (x->7) 2x / (x + 1).
A: Direct substitution gives 14/8 = 7/4 = 1.75.
Q: Evaluate lim (x->25) (x - 25) / (sqrt(x) - 5).
A: Direct substitution gives 0/0. Multiply numerator and denominator by (sqrt(x) + 5). The numerator becomes (x - 25)(sqrt(x) + 5) and the denominator becomes x - 25. Cancel (x - 25), leaving sqrt(x) + 5. Substitute x = 25 to get 10.
Q: Evaluate lim (x->infinity) (3x^2 - 7x + 1) / (6x^2 + 8).
A: Same degree (both quadratic). Ratio of leading coefficients: 3/6 = 1/2.
Q: Find the horizontal asymptote, vertical asymptote, and any holes of f(x) = (x - 5) / (x^2 - 4x - 5).
A: Factor the denominator: (x - 5)(x + 1). The (x - 5) cancels with the numerator, so there is a hole at x = 5 (y-coordinate: 1/(5 + 1) = 1/6). The remaining denominator gives a vertical asymptote at x = -1. Horizontal asymptote: the simplified function 1/(x + 1) has degree 0 on top and degree 1 on bottom, so HA is y = 0.
Q: Determine whether f(x) = {2x - 1 for x < 3, -5 for x = 3, x^2 - 4 for x > 3} is continuous at x = 3.
A: f(3) = -5 (defined). lim from left: 2(3) - 1 = 5. lim from right: 9 - 4 = 5. The two-sided limit is 5. But 5 does not equal -5, so f is not continuous at x = 3.
Q: Find the value of A that makes f(x) = {2x^2 - 3x + 1 for x < -2, Ax - 7 for x >= -2} continuous everywhere.
A: Set the pieces equal at x = -2. Left piece: 2(4) + 6 + 1 = 15. Right piece: -2A - 7. Solve -2A - 7 = 15, giving A = -11.
Limits are the foundation of derivatives. The definition of the derivative is itself a limit: f'(x) = lim (h->0) (f(x+h) - f(x)) / h. If you are comfortable evaluating limits, differentiation rules will feel like natural shortcuts.
Continuity connects to the Intermediate Value Theorem and to integration: a function must be continuous on [a, b] to guarantee the definite integral exists and behaves as expected.
Asymptote analysis feeds directly into curve sketching and graphing, which combines first and second derivative information with limit behaviour at the edges of the domain.
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