Difficulty: Introductory to Intermediate | Prerequisites: Algebra review (rewriting expressions in ax^n form, factoring, rationalising)
Limits are the foundation everything else in this course builds on: continuity, derivatives, and later optimisation all depend on understanding what a function approaches as its input gets close to some value. If you can evaluate limits confidently, the derivative rules that follow in the next unit will make sense rather than feeling like arbitrary formulas. You should already be comfortable with factoring, rationalising denominators, and rewriting expressions in the form ax^n.
A limit describes the value a function approaches as x gets close to a particular number (or heads towards infinity), even if the function never reaches that value. You evaluate most limits by direct substitution; when substitution gives 0/0, you factor, cancel, or rationalise first. For limits at infinity, the leading term of the numerator and denominator is all that matters.
Limit
The value that f(x) approaches as x gets closer and closer to a particular number c. Written lim (x→c) f(x). Think of it as: "Where is the function headed?" not "Where does it land?"
One-sided limit (left-hand and right-hand)
The value f(x) approaches as x comes from one direction only. lim (x→c⁻) is the left-hand limit (x approaches c from values less than c). lim (x→c⁺) is the right-hand limit (x approaches from values greater than c). In simple terms, you are only watching one side of the approach.
Two-sided limit
The ordinary limit lim (x→c) f(x). It exists only when the left-hand limit and right-hand limit are equal. If the two sides disagree, the two-sided limit Does Not Exist (DNE).
Limit at infinity
lim (x→∞) f(x) or lim (x→−∞) f(x). Describes the long-run behaviour of a function as x grows without bound in either direction. Think of it as: "What does the function settle down to eventually?"
Horizontal asymptote (HA)
A horizontal line y = L that the graph approaches as x → ∞ or x → −∞. You find it by evaluating the limit at infinity. In simple terms, it is the function's "ceiling" or "floor" far to the left or right.
Vertical asymptote (VA)
A vertical line x = a where the function blows up to ±∞. Occurs where the denominator equals zero and the numerator does not (the n/0 case). Think of it as an invisible wall the graph shoots along but never crosses.
Indeterminate form (0/0)
When direct substitution into a limit gives 0/0, the limit is not automatically zero or undefined. It means more work is needed: factor, cancel, or rationalise to find the true value. In simple terms, 0/0 is a "try again" signal, not an answer.
Piecewise function
A function defined by different rules on different intervals. When evaluating limits at the boundary between pieces, you must check the left-hand and right-hand limits separately, because each side uses a different formula.
Hole (removable discontinuity)
A point where the function is not defined (or defined at the wrong value), but the limit exists. On a graph it looks like a missing dot. You find holes when a 0/0 form cancels completely, leaving a finite limit at a point that is not in the domain.
If f(x) is a polynomial or a rational function and c is in the domain, just plug in: lim (x→c) f(x) = f(c).
Constants pass through unchanged: lim (x→c) k = k, regardless of what c is.
Example: lim (x→3) 17 = 17. The constant does not care what x does.
For a sum or product, evaluate each piece and combine: lim (x→5) (x² + 3x + 2) = 25 + 15 + 2 = 42.
For a quotient, substitute top and bottom separately: lim (x→−5) (x²+3)/(x+4) = 28/(−1) = −28.
A piecewise function uses different formulas on different intervals. At the boundary, the left-hand limit comes from the left piece, the right-hand limit from the right piece.
The two-sided limit exists only if both one-sided limits are equal.
Example: f(x) = { x+1 if x < −1; 3−x if x ≥ −1 }. At x = −1:
Left: lim (x→−1⁻) (x+1) = 0
Right: lim (x→−1⁺) (3−x) = 4
Since 0 ≠ 4, lim (x→−1) f(x) = DNE. But f(−1) = 3−(−1) = 4 (uses the ≥ piece).
When both sides agree, the limit exists and equals that shared value.
Example: f(x) = { x+1 if x < 3; 10−2x if x ≥ 3 }. At x = 3:
Left: 3+1 = 4. Right: 10−6 = 4. Both give 4, so lim (x→3) f(x) = 4.
When substitution gives 0/0, the limit is not dead. Use algebra to simplify, then substitute again.
Factor and cancel: lim (x→2) (x²−4)/(x−2) = lim (x→2) (x+2)(x−2)/(x−2) = lim (x→2) (x+2) = 4.
Rationalise the numerator: lim (x→16) (√x−4)/(x−16). Multiply top and bottom by (√x+4): the numerator becomes x−16, which cancels with the denominator. Result: lim (x→16) 1/(√x+4) = 1/8.
Rationalise the denominator: lim (x→100) (x−100)/(√x−10). Multiply by (√x+10)/(√x+10): the denominator becomes x−100, which cancels. Result: lim (x→100) (√x+10) = 20.
Difference of squares for higher powers: lim (x→8) (x²−64)/(x−8) = lim (x→8) (x+8)(x−8)/(x−8) = 16.
For a rational function, divide every term by the highest power of x in the denominator and let x → ∞.
Shortcut (leading-term test): compare the degree of the numerator (n) to the degree of the denominator (d).
n < d: the limit is 0. The HA is y = 0.
n = d: the limit is the ratio of leading coefficients. Example: lim (x→∞) (4x−7)/(2x+3) = 4/2 = 2.
n > d: no horizontal asymptote. The function grows without bound.
For polynomials, the leading term determines the end behaviour.
Example: f(x) = 2x² − 5x + 8. Leading term is 2x². Since the coefficient is positive and the power is even, lim (x→±∞) f(x) = ∞.
Example: f(x) = 9x⁷ − 5x⁴ − x² + 8. Leading term is 9x⁷. Odd power with positive coefficient: lim (x→∞) f(x) = ∞, lim (x→−∞) f(x) = −∞.
When substitution gives n/0 (a nonzero number over zero), the function blows up. The sign depends on the direction of approach.
Example: lim (x→2⁻) 5/(x−2). As x → 2 from the left, (x−2) is a small negative number, so 5/(small negative) → −∞.
Example: lim (x→2⁻) x³/(x−2)². The denominator (x−2)² is always positive, the numerator approaches 8, so the limit is +∞.
Set the denominator equal to zero. At each such value of x, substitute into the original to check:
If you get n/0, there is a vertical asymptote.
If you get 0/0, try factoring. The cancelled factor usually gives a hole.
The y-coordinate of a hole is the value of the simplified function at that x.
Example: f(x) = (x−1)/(x²−1) = (x−1)/((x−1)(x+1)). At x = 1: 0/0, so factor and cancel to get 1/(x+1). The hole is at (1, 1/2). At x = −1: −2/0 = n/0, so x = −1 is a VA.
The horizontal asymptote comes from the limit at ±∞ using the leading-term test.
Business production models often ask what happens "in the long run," which is a limit at infinity question. For example, if P(t) = (8t²+5t)/(2(t+1)²) thousand units, then lim (t→∞) P(t) = 8/2 = 4, telling you production levels off at 4,000 units.
Constant rule: lim (x→c) k = k
Identity rule: lim (x→c) x = c
Sum/difference rule: lim (x→c) [f(x) ± g(x)] = lim f(x) ± lim g(x)
Product rule: lim (x→c) [f(x) · g(x)] = lim f(x) · lim g(x)
Quotient rule: lim (x→c) f(x)/g(x) = lim f(x) / lim g(x), provided lim g(x) ≠ 0
Power rule: lim (x→c) [f(x)]^n = [lim f(x)]^n
Root rule: lim (x→c) √[f(x)] = √[lim f(x)], provided the result is real
Leading-term test for rational functions at infinity:
Given f(x) = (a_n x^n + …) / (b_d x^d + …):
If n < d: lim = 0
If n = d: lim = a_n / b_d
If n > d: lim = ±∞ (no HA)
Two-sided limit existence condition:
lim (x→c) f(x) = L if and only if lim (x→c⁻) f(x) = L and lim (x→c⁺) f(x) = L.
Students often think that if f(c) exists, then lim (x→c) f(x) must equal f(c). It does not have to. The limit is about where the function is heading, not where it actually lands. These two values are equal only when the function is continuous at c.
Students often think 0/0 means the limit is zero or undefined. It does not. 0/0 is an indeterminate form, which means you need to do more algebra. The actual limit could be any finite number.
Students often confuse "the limit is ∞" with "the limit exists." When we write lim = ∞, we are saying the function grows without bound. Technically, the limit does not exist as a finite number. However, it is still useful information (it tells you there is a vertical asymptote).
Students often forget to check both one-sided limits for piecewise functions. If you only check one side, you can miss a DNE situation. Always check left and right at the boundary.
⚠️ Reading limits from a graph (one-sided, two-sided, at infinity) is a near-certain exam question. Know how to read open circles vs filled dots.
⚠️ 0/0 indeterminate forms requiring factoring or rationalising are tested with full working required. Showing your algebra step-by-step is worth marks.
⚠️ Piecewise limit problems always test whether the two-sided limit exists by comparing left and right sides. The exam also asks for f(c) separately.
⚠️ Limits at infinity and identifying horizontal/vertical asymptotes and holes will appear. You must use limit notation to receive credit, not just state the answer.
⚠️ The long-run production/business application (limit at infinity of a rational function) is a standard word problem on this exam.
True or False: If f(3) = 7, then lim (x→3) f(x) must also be 7. False. The function value and the limit are independent unless the function is continuous there.
Fill in the blank: When direct substitution into a limit gives 0/0, this is called an ______ form. Indeterminate.
True or False: lim (x→∞) (3x² + 1)/(5x² − 2) = 3/5. True. Same degree, so take the ratio of leading coefficients.
Fill in the blank: A two-sided limit exists if and only if the ______ limit equals the ______ limit. Left-hand; right-hand.
True or False: If a rational function gives n/0 at x = a, there is a hole at x = a. False. n/0 indicates a vertical asymptote, not a hole. Holes come from 0/0 forms that cancel.
Q: Evaluate lim (x→2) (x² − 4)/(x − 2). Show your work.
A: Direct substitution gives 0/0. Factor the numerator: (x+2)(x−2)/(x−2). Cancel (x−2). You get lim (x→2) (x+2) = 4.
Q: For f(x) = { x+1 if x < −1; 3−x if x ≥ −1 }, find lim (x→−1) f(x) and f(−1).
A: Left: lim (x→−1⁻) (x+1) = 0. Right: lim (x→−1⁺) (3−x) = 4. Since 0 ≠ 4, the limit DNE. f(−1) = 3−(−1) = 4.
Q: Find all vertical asymptotes, horizontal asymptotes, and holes for f(x) = (x−1)/(x²−1).
A: Factor denominator: (x−1)(x+1). At x = 1: gives 0/0, cancel to get 1/(x+1), so there is a hole at (1, 1/2). At x = −1: gives −2/0 = n/0, so VA at x = −1. For the HA: degree of numerator (1) < degree of denominator (2), so HA is y = 0.
Q: Evaluate lim (x→∞) (4x − 7)/(2x + 3).
A: Same degree (both degree 1). Ratio of leading coefficients: 4/2 = 2. The limit is 2.
Q: A business manager models production as P(t) = (8t² + 5t) / (2(t+1)²) thousand units. What happens in the long run?
A: Expand the denominator: 2(t² + 2t + 1) = 2t² + 4t + 2. Leading terms: 8t² / 2t² = 4. Production approaches 4,000 units.
Limits feed directly into continuity (the next section of this exam): a function is continuous at a point only if its limit there equals its function value. Limits also define the derivative, which is the core of the rest of this course. Every differentiation problem you do later is, at its foundation, a limit of a difference quotient.
Limit, one-sided limit, left-hand limit, right-hand limit, two-sided limit, limit at infinity, end behaviour, horizontal asymptote, vertical asymptote, hole, removable discontinuity, indeterminate form, 0/0, direct substitution, factor and cancel, rationalise, rationalizing, leading-term test, degree comparison, piecewise function, long-run behaviour, MAC2233, calculus for business, FIU, exam 1 review, sections 2.1 2.2 2.3