Difficulty: Introductory to Intermediate. Prerequisites: college algebra, familiarity with polynomial and rational functions.
A limit describes what value a function's output approaches as the input gets close to some number, even if the function never reaches that value. If the left-hand and right-hand limits agree, the two-sided limit exists. Continuity means the function has no holes, jumps, or breaks at a given point, which requires the limit to exist and equal the function's value there.
Limit
The value that f(x) approaches as x gets closer and closer to some number c from both sides. Written as lim(x→c) f(x) = L.
In simple terms, this means: "What y-value is the graph heading towards as x closes in on c?"
One-sided limit (left-hand limit)
The value f(x) approaches as x tends toward c from values less than c (from the left). Written as lim(x→c⁻) f(x) = L.
Think of it as: "What is the y-value doing as I walk along the graph towards c from the left side only?"
One-sided limit (right-hand limit)
The value f(x) approaches as x tends toward c from values greater than c (from the right). Written as lim(x→c⁺) f(x) = M.
Think of it as: "What is the y-value doing as I walk along the graph towards c from the right side only?"
Indeterminate form (0/0)
The result you get when direct substitution into a limit yields 0/0. This does not mean the limit is zero or undefined. It means the expression can be simplified, typically by factoring or multiplying by the conjugate, and then re-evaluated.
In simple terms, this means: "0/0 is a signal to simplify, not an answer."
Horizontal asymptote
A horizontal line y = L that the graph of a function approaches as x goes to positive or negative infinity. Found by evaluating lim(x→±∞) f(x). If that limit is a finite number, that number is the horizontal asymptote.
Think of it as: "Where does the graph flatten out in the long run?"
Vertical asymptote
A vertical line x = c where the function's output grows without bound (towards ∞ or −∞). Occurs when plugging in x = c gives a nonzero number divided by zero (n/0, where n ≠ 0).
Think of it as: "Where does the graph shoot off to infinity because the denominator hits zero and the numerator does not?"
Hole (removable discontinuity)
A point where the function is not defined (or is defined at a different value) but the limit exists. Graphically, it looks like a gap or open circle on an otherwise continuous curve. Occurs when both numerator and denominator share a common factor that cancels (0/0 after substitution).
In simple terms, this means: "The graph has a tiny missing dot, not a full break."
Continuity at a point
A function f(x) is continuous at x = c if three conditions hold: (I) f(c) is defined, (II) lim(x→c) f(x) exists, and (III) lim(x→c) f(x) = f(c).
Think of it as: "You can draw through the point without lifting your pencil."
End behaviour
The behaviour of f(x) as x approaches positive or negative infinity. For polynomials, the leading term determines whether the output goes to ∞ or −∞.
In simple terms, this means: "What does the far left and far right of the graph do?"
To find lim(x→c) f(x) from a graph, look at where the y-values are heading as x approaches c from both sides.
The actual value of f(c), or whether f(c) even exists, does not matter for the limit. The limit is about the trend, not the destination.
If the graph approaches the same y-value from the left and right, that y-value is the limit.
The left-hand limit lim(x→c⁻) f(x) = L looks at x-values less than c approaching c.
The right-hand limit lim(x→c⁺) f(x) = M looks at x-values greater than c approaching c.
The two-sided limit exists if and only if the left-hand limit equals the right-hand limit. If L ≠ M, the two-sided limit does not exist (DNE).
Example: For f(x) = {1 − x² if 0 ≤ x < 2, 2x + 1 if x ≥ 2}, lim(x→2⁻) f(x) = 1 − 4 = −3 and lim(x→2⁺) f(x) = 5. Since −3 ≠ 5, the two-sided limit at x = 2 does not exist.
Step 1: Try direct substitution. Plug c into f(x). If you get a real number, that is the limit.
For polynomials: lim(x→c) p(x) = p(c). Always works.
For rational functions: lim(x→c) p(x)/q(x) = p(c)/q(c), provided q(c) ≠ 0.
Step 2: If you get 0/0 (indeterminate form), simplify the function.
Factor and cancel common factors, then re-substitute.
If the expression has a square root, multiply numerator and denominator by the conjugate, then re-substitute.
Example (factoring): lim(x→1) (x² − 1)/(x² − 3x + 2) = lim(x→1) (x+1)(x−1)/((x−1)(x−2)) = lim(x→1) (x+1)/(x−2) = 2/(−1) = −2.
Example (conjugate): lim(x→1) (√x − 1)/(x − 1). Multiply by (√x + 1)/(√x + 1) to get 1/(√x + 1). Substituting gives 1/2.
For any polynomial f(x), as x → ±∞, the output also goes to ±∞. The leading term determines the sign.
Even-degree leading term with positive coefficient: f(x) → ∞ as x → ∞ and x → −∞ (both ends up).
Even-degree leading term with negative coefficient: f(x) → −∞ in both directions (both ends down).
Odd-degree leading term with positive coefficient: f(x) → −∞ as x → −∞, and f(x) → ∞ as x → ∞ (down-left, up-right).
Odd-degree leading term with negative coefficient: f(x) → ∞ as x → −∞, and f(x) → −∞ as x → ∞ (up-left, down-right).
Example: lim(x→∞) (3x² − x + 1) = ∞ because the leading term 3x² has even power and positive coefficient.
Compare the degree of the numerator (top) to the degree of the denominator (bottom).
Same degree: the limit equals the ratio of the leading coefficients.
Example: lim(x→∞) (2x² + 3x + 1)/(3x² − 5x + 2) = 2/3.
Numerator degree less than denominator degree: the limit is 0.
Example: lim(x→−∞) (3x³ − 6)/(x⁴ − 5x² + 1) = 0.
Numerator degree greater than denominator degree: the limit is ±∞ (no horizontal asymptote).
Example: lim(x→−∞) (7x⁶ + 3)/(x⁴ + x²) = ∞.
Reciprocal limit (very important): lim(x→±∞) k/x^p = 0 for any constant k and positive power p.
Horizontal asymptote: Take lim(x→±∞) f(x). If the result is a finite number L, then y = L is a horizontal asymptote.
Vertical asymptote: Find the x-values that make the denominator zero. Substitute into the function.
If you get n/0 (nonzero over zero), there is a vertical asymptote at that x-value.
If you get 0/0, factor and cancel. The cancelled factor gives a hole, not an asymptote. Evaluate the simplified function at that x to find the hole's y-coordinate.
Example: f(x) = (x − 5)/(x² − 3x − 10). The denominator factors as (x − 5)(x + 2). The (x − 5) cancels, giving a hole at x = 5 (with y-coordinate 1/7). The remaining factor (x + 2) = 0 gives a vertical asymptote at x = −2.
A function f(x) is continuous at x = c if all three conditions hold:
(I) f(c) is defined.
(II) lim(x→c) f(x) exists (left-hand limit = right-hand limit).
(III) lim(x→c) f(x) = f(c) (the limit equals the function value).
Polynomials are continuous on all real numbers, always.
Rational functions are continuous everywhere except where the denominator equals zero.
For piecewise functions, the only potential trouble spot is where the rule changes (the boundary value).
Check all three continuity conditions at the boundary.
Example: f(x) = {x + 1 if x ≤ 2, 2 if x > 2}. At x = 2: f(2) = 3, but lim(x→2⁺) f(x) = 2. Since 3 ≠ 2, the function is not continuous at x = 2.
Finding the constant for continuity: If f(x) = {Ax + 5 if x < 1, x² − 3x + 4 if x ≥ 1}, set the left-hand limit at x = 1 equal to the right-hand limit: A(1) + 5 = (1)² − 3(1) + 4, so A + 5 = 2, giving A = −3.
\lim_{x \to c} k = k \quad \text{(constant rule)}\lim_{x \to c} x = c\lim_{x \to c} p(x) = p(c) \quad \text{(polynomial direct substitution)}\lim_{x \to c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}, \quad q(c) \neq 0\lim_{x \to \pm\infty} \frac{k}{x^p} = 0 \quad \text{(reciprocal limit, } p > 0\text{)}Continuity checklist at x = c:
f(c) is defined.
lim(x→c) f(x) exists (left limit = right limit).
lim(x→c) f(x) = f(c).
Asymptote-finding steps for rational functions:
Horizontal: evaluate lim(x→±∞) f(x) using leading-term comparison.
Vertical: set denominator = 0, then test each value. If substitution gives n/0 (n ≠ 0), it is a vertical asymptote. If it gives 0/0, factor and cancel to find a hole.
Limits underpin how businesses model change. When an economist says "as production increases without bound, the average cost approaches $10 per unit," that is a limit at infinity in action, and the $10 figure is a horizontal asymptote of the average cost function. Continuity matters in pricing models and tax brackets: a discontinuous tax function means a tiny income increase could cause a disproportionate jump in tax owed. In finance, continuous compounding (covered later in this course) is defined as the limit of the compound interest formula as the number of compounding periods goes to infinity.
Students often think the limit of f(x) at x = c must equal f(c). It does not. The limit depends on what the function approaches, not on the function's actual value at c. A function can have a limit of 3 at x = 2 even if f(2) = 7, or if f(2) is undefined.
Students often think that 0/0 means the limit is zero or does not exist. 0/0 is the indeterminate form, which is a prompt to simplify. The limit could be any finite number.
Students often confuse "the limit does not exist" with "the limit is infinity." These are different statements. A limit of ∞ means the function grows without bound (and technically does not exist as a finite value, but it tells you the behaviour). "Does not exist" with no further detail usually means the left-hand and right-hand limits disagree.
Students often forget to check all three conditions for continuity. Showing that the limit exists is not enough; you must also verify that f(c) is defined and that the limit equals f(c).
⚠️ Evaluating a limit algebraically: the first step is always to try direct substitution. If that gives 0/0, factor or use the conjugate. This sequence appears on nearly every exam.
⚠️ Knowing when a two-sided limit does not exist: if the left-hand and right-hand limits give different values, write DNE and state both one-sided limits.
⚠️ Rational function limits at infinity: you are expected to determine horizontal asymptotes quickly by comparing leading-term degrees. Same degree = ratio of leading coefficients, smaller numerator degree = 0, larger numerator degree = no horizontal asymptote.
⚠️ Distinguishing holes from vertical asymptotes: 0/0 after substitution means a hole (factor and cancel), n/0 means a vertical asymptote. Exam questions will ask you to identify which is which and to give the coordinates of any hole.
⚠️ Continuity of piecewise functions: expect a question asking for the value of a constant that makes a piecewise function continuous at the boundary. Set the left-hand limit equal to the right-hand limit and solve.
True or false: If f(3) = 5, then lim(x→3) f(x) = 5.
Fill in the blank: If direct substitution into a limit gives 0/0, this is called the __________ form.
True or false: A rational function whose numerator has a lower degree than its denominator has a horizontal asymptote at y = 0.
Fill in the blank: For lim(x→c) f(x) to exist, the __________ limit must equal the __________ limit.
True or false: If a function has a hole at x = 4, it also has a vertical asymptote at x = 4.
Answers: 1. False (the limit depends on the approach, not the value). 2. Indeterminate. 3. True. 4. Left-hand, right-hand. 5. False (a hole and a vertical asymptote are mutually exclusive at the same x-value).
Q: Evaluate lim(x→−1) (3x² − 4x + 8).
A: Direct substitution. 3(−1)² − 4(−1) + 8 = 3 + 4 + 8 = 15.
Q: Evaluate lim(x→1) (x² − 1)/(x² − 3x + 2).
A: Substitution gives 0/0. Factor: (x+1)(x−1) / (x−1)(x−2). Cancel (x−1). Result: (x+1)/(x−2). Substitute x = 1: 2/(−1) = −2.
Q: Find the horizontal and vertical asymptotes of f(x) = (2x² + 4x)/(x² − 1).
A: Horizontal: degrees are equal, so y = 2/1 = 2. Vertical: x² − 1 = 0 gives x = 1 and x = −1. Check for holes: substituting x = 1 gives 6/0 (vertical asymptote). Substituting x = −1 gives −2/0 (vertical asymptote). Both are vertical asymptotes; no holes.
Q: Determine whether g(x) = (x² − 1)/(x − 1) is continuous at x = 1.
A: g(1) is undefined (0/0), so condition (I) fails immediately. The function is not continuous at x = 1. (Note: the limit exists and equals 2, since g simplifies to x + 1 for x ≠ 1, but continuity still fails because g(1) is not defined.)
Q: For what value of A is f(x) = {Ax + 5 if x < 1, x² − 3x + 4 if x ≥ 1} continuous for all real x?
A: Set left-hand limit at x = 1 equal to right-hand limit. Left: A(1) + 5 = A + 5. Right: 1 − 3 + 4 = 2. So A + 5 = 2, giving A = −3.
Q: Evaluate lim(x→∞) (2x² + 3x + 1)/(3x² − 5x + 2).
A: Same degree (both degree 2). Limit = ratio of leading coefficients = 2/3.
Limits are the foundation of everything that follows in this course. The derivative is defined as a specific limit (the limit of the difference quotient as h → 0), so a solid grasp of limit evaluation is essential before moving to differentiation. Continuity connects directly to the Intermediate Value Theorem and to understanding where derivatives exist, since a function must be continuous at a point to be differentiable there. Asymptotic behaviour reappears when you analyse cost, revenue, and profit functions as production scales up.
Tags: limits, one-sided limits, left-hand limit, right-hand limit, indeterminate form, 0/0, factoring limits, conjugate method, limits at infinity, end behaviour, horizontal asymptote, vertical asymptote, hole, removable discontinuity, continuity, piecewise continuity, three conditions for continuity, finding the constant for continuity, reciprocal limits, polynomial limits, rational function limits, MAC 2233, business calculus, FIU