Difficulty: Introductory to Intermediate | Prerequisites: Algebra (factoring, rationalising, polynomial long division)
Limits are the foundation everything else in calculus rests on. Before you can talk about derivatives (rates of change) or integrals (accumulated totals), you need a way to describe what a function's output approaches as the input gets close to some value. This chapter introduces that language, extends it to behaviour at infinity (horizontal asymptotes), and then uses limits to define continuity, the property that lets you evaluate a limit just by plugging in. If you skipped college algebra's section on factoring and rationalising, go back to it first: those two techniques do most of the heavy lifting here.
A limit describes the value a function approaches as x gets close to a target, even if the function is undefined there. For polynomials and rational functions you can often just substitute; when substitution gives 0/0, factor or rationalise first. Limits at infinity tell you what happens as x grows without bound, and the three continuity conditions let you test whether a function has any breaks.
Limit
The value that f(x) approaches as x approaches a particular number c. Written lim (x→c) f(x) = L. The function does not need to equal L at c, and it does not even need to be defined at c. Think of it as: "Where is the function headed?" not "Where is it right now?"
One-sided limit (left-hand and right-hand)
The value f(x) approaches as x comes from one direction only. lim (x→c⁻) f(x) is the left-hand limit (x approaches c from below). lim (x→c⁺) f(x) is the right-hand limit (x approaches c from above). In simple terms, you are only looking at the function's behaviour on one side of the target.
Two-sided limit
The ordinary limit lim (x→c) f(x). It exists only when both one-sided limits exist and are equal. If the left and right sides disagree, the two-sided limit does not exist (DNE).
Continuity at a point
A function f is continuous at x = c when three conditions hold simultaneously: f(c) is defined, lim (x→c) f(x) exists, and lim (x→c) f(x) = f(c). In simple terms, you can draw through that point without lifting your pen.
Polynomial function
A function built from whole-number powers of x with constant coefficients, such as f(x) = 3x⁴ – 5x + 7. No fractions with x in the denominator, no roots of x, no other operations. Polynomials are continuous everywhere.
Rational function
A ratio of two polynomials, such as g(x) = (x² – 1)/(x + 3). Continuous everywhere except where the denominator equals zero.
Indeterminate form (0/0)
What you get when direct substitution into a rational expression gives 0 in both numerator and denominator. This does not mean the limit is zero or undefined; it means more work is needed (factor, rationalise, or simplify).
Vertical asymptote
A vertical line x = a where the function grows without bound. Found where the denominator of a simplified rational function equals zero. The limit from at least one side is +∞ or –∞.
Horizontal asymptote
A horizontal line y = L that the function approaches as x → ∞ or x → –∞. Found by evaluating limits at infinity.
Hole (removable discontinuity)
A point where both numerator and denominator are zero, so a common factor cancels. The limit exists there, but the function is undefined. Plotted as an open circle on the graph.
For any polynomial, just plug in. lim (x→2) (3x² – x) = 3(2)² – 2 = 10.
For a rational function, plug in provided the denominator is not zero. lim (x→–2) (x+5)/(2x–3) = (–2+5)/(2(–2)–3) = 3/(–7) = –3/7.
The limit of a constant is that constant. lim (x→6) (–8) = –8, regardless of what x does.
Factor and cancel. If substitution gives 0/0, factor numerator and denominator, cancel the common factor, then substitute again.
Example: lim (x→–1) (2x²–x–3)/(x²–1). Both top and bottom equal 0 at x = –1. Factor: (2x–3)(x+1) / (x–1)(x+1). Cancel (x+1): (2x–3)/(x–1). Now substitute: (2(–1)–3)/((–1)–1) = –5/–2 = 5/2.
Rationalise. When a square root appears, multiply top and bottom by the conjugate.
Example: lim (x→25) (x–25)/(√x–5). Multiply by (√x+5)/(√x+5): numerator becomes (x–25)(√x+5), denominator becomes x–25. Cancel: √x+5. Substitute: √25+5 = 10.
For a polynomial, the highest-degree term dominates.
lim (x→∞) (–5x⁶ + 5x + 1) = –∞ (leading term –5x⁶ → –∞).
lim (x→–∞) (3x³ – 4x² + x) = –∞ (odd power, negative direction: 3(–∞)³ = –∞).
For a rational function, compare the degree of the numerator (N) with the degree of the denominator (D).
N < D: limit is 0.
N = D: limit is the ratio of leading coefficients.
N > D: limit is ±∞ (no horizontal asymptote).
Example: lim (x→∞) (3–x–4x²)/(3x²–1). N = D = 2. Ratio of leading coefficients: –4/3.
A left-hand limit lim (x→c⁻) uses the piece of the function valid for x < c.
A right-hand limit lim (x→c⁺) uses the piece valid for x > c.
The two-sided limit exists only when both sides agree.
The function value f(c) uses whichever piece is defined at x = c (which may differ from both one-sided limits).
Example with f(x) = {4–x² for x<3, 2x+7 for x=3, 9 for x>3}:
lim (x→3⁻) f(x) = 4–9 = –5 (use 4–x²).
lim (x→3⁺) f(x) = 9 (use 9).
lim (x→3) f(x) DNE (left ≠ right).
f(3) = 2(3)+7 = 13 (use the x=3 piece).
Factor both numerator and denominator of the rational function.
Any factor that cancels produces a hole at that x-value. The y-coordinate of the hole is found by substituting into the simplified function.
Any factor left in the denominator after cancelling produces a vertical asymptote at that x-value.
Example: f(x) = (x²+x–6)/(x²–4) = (x+3)(x–2)/((x+2)(x–2)). The (x–2) cancels, so there is a hole at x = 2. Substituting into (x+3)/(x+2) gives y = 5/4, so the hole is at (2, 5/4). The factor (x+2) remains, so x = –2 is a vertical asymptote.
Three conditions must all hold at x = c:
f(c) is defined.
lim (x→c) f(x) exists.
lim (x→c) f(x) = f(c).
Polynomials are continuous everywhere.
Rational functions are continuous on their domain (everywhere except where the denominator is zero).
For piecewise functions, check continuity at the boundary values by verifying all three conditions.
Polynomial (P): Only whole-number powers of x. Examples: f(x) = 3x⁴ – 5x + 7, f(x) = (3x–1)(2x+5) (multiply out to get a polynomial).
Rational (R): A ratio of two polynomials. Examples: g(x) = 3/(x–1), m(x) = (x²–2x–1)/(x³+2).
Neither (N): Contains roots of x, absolute values, or other non-polynomial operations. Examples: h(x) = ³√x – 1/x + 1, f(x) = (√x–4)/(x+1).
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
For f(x) = (a_n x^n + ...)/(b_m x^m + ...) as x → ±∞:
If n < m: lim = 0 (horizontal asymptote y = 0)
If n = m: lim = a_n / b_m (horizontal asymptote y = a_n / b_m)
If n > m: lim = ±∞ (no horizontal asymptote)
f(c) is defined (the point exists).
lim (x→c) f(x) exists (left and right limits agree).
lim (x→c) f(x) = f(c) (the limit equals the function value).
Students often write the limit notation on every step of the algebra. Once you substitute, the limit symbol should be dropped. Writing "lim (x→2) 10" as a final answer is incorrect; the answer is simply 10.
Students often confuse "the limit does not exist" with "the function is undefined." A limit can exist even when the function is undefined at that point (that is exactly what happens at a hole).
Students often think that 0/0 means the limit is 0 or is undefined. 0/0 is an indeterminate form, which just means more algebraic work is required before the limit can be determined.
Students often forget that the value of a piecewise function at the boundary, f(c), uses the piece defined at x = c, which can be completely different from the left-hand or right-hand limits.
⚠️ Proper limit notation is required for full credit. Drop the lim symbol only after you have finished substituting.
⚠️ Expect at least one 0/0 problem that requires factoring or rationalising before evaluating.
⚠️ Piecewise limit problems will ask for one-sided limits, the two-sided limit, and the function value separately. These can all be different numbers.
⚠️ You may be asked to identify two properties of a rational function at a point where it is undefined (e.g. the limit exists but the function has a hole).
⚠️ Finding the value of A that makes a piecewise function continuous is a standard exam question. Set the left-hand limit equal to the function value and solve.
⚠️ Asymptotes must be found using limits, not by inspection. Show the limit statement.
True or False: lim (x→5) 12 = 12. (True. The limit of a constant is that constant.)
True or False: If lim (x→c) f(x) exists but f(c) is undefined, then f is continuous at c. (False. Condition 1 of continuity fails.)
Fill in the blank: When direct substitution into a rational function gives 0/0, the result is called an __________ form. (indeterminate)
True or False: For f(x) = (3x⁴ + 2)/(5x⁴ – 1), the horizontal asymptote is y = 3/5. (True. Same degree, ratio of leading coefficients.)
Fill in the blank: If the left-hand limit is 4 and the right-hand limit is 7, the two-sided limit __________. (does not exist / DNE)
Q: Evaluate lim (x→–1) (2x² – x – 3)/(x² – 1).
A: Factor: (2x–3)(x+1)/((x–1)(x+1)). Cancel (x+1): (2x–3)/(x–1). Substitute x = –1: (–2–3)/(–1–1) = –5/–2 = 5/2.
Q: Evaluate lim (x→25) (x–25)/(√x – 5).
A: Multiply by (√x+5)/(√x+5). Numerator becomes (x–25)(√x+5). Denominator becomes x–25. Cancel: result is √x+5. Substitute x = 25: 5+5 = 10.
Q: Evaluate lim (x→∞) (–2x⁵)/((x+1)²).
A: Expand the denominator: (x+1)² = x² + 2x + 1, degree 2. Numerator degree is 5, which is greater. Since the leading term is –2x⁵/x² = –2x³ → –∞, the limit is –∞.
Q: For f(x) = (x²+x–6)/(x²–4), find all vertical asymptotes, horizontal asymptotes, and holes.
A: Factor: (x+3)(x–2)/((x+2)(x–2)). Cancel (x–2): hole at x = 2, y-coordinate = (2+3)/(2+2) = 5/4. Vertical asymptote at x = –2 (remaining denominator factor). Horizontal asymptote: degrees equal, so y = 1/1 = 1.
Q: Determine the value of A so that f(x) = {–5 for x < –1, x–4 for x = –1, x²+Ax for x > –1} is continuous everywhere.
A: For continuity at x = –1, the limit from both sides must equal f(–1). f(–1) = –1–4 = –5. Left-hand limit = –5. Right-hand limit = (–1)² + A(–1) = 1 – A. Set 1 – A = –5, so A = 6.
Q: Is f(x) = (3x–1)(2x+5) a polynomial, rational, or neither?
A: Polynomial (P). Multiplying out gives 6x² + 13x – 5, which contains only whole-number powers of x.
Limits are the formal machinery behind derivatives (Section 2.4 defines the derivative as a limit) and integrals (definite integrals are limits of Riemann sums). Continuity matters because the Extreme Value Theorem, which guarantees absolute extrema on a closed interval, requires the function to be continuous. Asymptotes reappear in the graphing strategy sections (4.1–4.4), where you sketch rational functions using limits, intercepts, and derivative sign charts together.
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