Difficulty: Foundational. Prerequisites: basic algebra, factoring, rationalisation, familiarity with function notation.
Limits describe what value a function approaches as x gets close to some number (or heads to infinity), even if the function never reaches that value. Continuity means a function has no breaks, jumps, or holes at a point. Together these two ideas underpin every derivative and integral you will meet in the rest of the course.
Limit
The value a function f(x) approaches as x approaches a particular number c. Written lim (x→c) f(x) = L. The function does not have to equal L at c; it only has to get arbitrarily close.
In simple terms: "Where is the y-value heading as x sneaks up on c?"
One-sided limit (left-hand)
The value f(x) approaches as x approaches c from the left (from values smaller than c). Written lim (x→c⁻) f(x).
Think of it as: approaching c on the number line from the left side only.
One-sided limit (right-hand)
The value f(x) approaches as x approaches c from the right (from values larger than c). Written lim (x→c⁺) f(x).
Think of it as: approaching c on the number line from the right side only.
Two-sided limit
The overall limit lim (x→c) f(x) exists only when the left-hand and right-hand limits both exist and are equal.
In simple terms: the function must be heading to the same y-value from both directions.
Indeterminate form
An expression such as 0/0 that arises from direct substitution and tells you nothing about the actual limit. You must simplify (factor, rationalise, etc.) before evaluating.
Think of it as: a "try again" signal, not an answer.
Vertical asymptote
A vertical line x = a where the function grows without bound (toward +∞ or −∞) as x approaches a. The denominator equals zero at x = a and the numerator does not.
In simple terms: the graph shoots up or down toward a wall it never crosses.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x → +∞ or x → −∞. Found by comparing the degrees of numerator and denominator in a rational function.
Think of it as: the long-run levelling-off value of the function.
Continuity at a point
A function f is continuous at x = c when three conditions hold: (1) f(c) is defined, (2) lim (x→c) f(x) exists, and (3) lim (x→c) f(x) = f(c).
In simple terms: you can draw through the point without lifting your pen.
Piecewise function
A function defined by different rules on different intervals. Continuity questions on exams almost always involve piecewise functions, because the potential break is at the boundary between pieces.
Trace along the curve from the left toward the target x-value; the y-value you approach is the left-hand limit.
Trace from the right; the y-value you approach is the right-hand limit.
If both sides agree, that shared value is the two-sided limit.
An open circle means the function is not defined there (or takes a different value), but the limit can still exist.
A filled dot at a different y-value means f(c) exists but differs from the limit, so the function is not continuous.
Always try plugging c into f(x) first. If you get a real number, that number is the limit. This works whenever f is continuous at c (polynomials, exponentials, basic rationals with non-zero denominators).
Example: lim (x→5) (x² + 3x + 2) = 25 + 15 + 2 = 42.
When direct substitution gives 0/0, simplify and try again.
Factoring: Factor numerator and denominator, cancel the common factor, then substitute.
Example: lim (x→2) (x² − 4)/(x − 2) = lim (x→2) (x − 2)(x + 2)/(x − 2) = lim (x→2) (x + 2) = 4.
Rationalising: Multiply top and bottom by the conjugate when a square root is involved.
Example: lim (x→16) (√x − 4)/(x − 16). Multiply by (√x + 4)/(√x + 4) to get 1/(√x + 4). Substituting gives 1/8.
Expanding and simplifying: Sometimes expanding a product and cancelling is the path.
Example: lim (x→100) (x − 100)/(√x − 10). Rationalise to get √x + 10. Result = 20.
For a piecewise function, use the rule that applies on the side you are approaching from.
Left-hand limit at the boundary: use the formula for x < boundary.
Right-hand limit at the boundary: use the formula for x ≥ boundary (or x > boundary).
If left ≠ right, the two-sided limit does not exist (DNE).
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 equal 4, so the limit exists and equals 4.
For a rational function p(x)/q(x):
Degree of numerator < degree of denominator: limit = 0.
Degrees equal: limit = ratio of leading coefficients.
Degree of numerator > degree of denominator: limit = ±∞ (no horizontal asymptote).
Example: lim (x→∞) (4x − 7)/(2x + 3) = 4/2 = 2.
When the denominator approaches 0 and the numerator does not, the function blows up to +∞ or −∞.
Check the sign on each side to determine +∞ or −∞.
Vertical asymptote at every x-value that makes the denominator zero (after cancellation).
Example: lim (x→2⁻) 5/(x − 2). As x → 2 from the left, (x − 2) is small and negative, so 5/(x − 2) → −∞.
Vertical: Set denominator = 0 and solve. Check that the factor does not cancel.
Horizontal: Compare degrees (rules above). If degree of numerator = degree of denominator + 1, there is an oblique asymptote instead (not tested in this section).
Example: f(x) = 2x/(x + 2). Vertical asymptote at x = −2. Horizontal asymptote: degrees equal, so y = 2/1 = 2.
A function f is continuous at x = c if and only if all three conditions hold:
f(c) is defined (the point exists on the graph).
lim (x→c) f(x) exists (left and right limits agree).
lim (x→c) f(x) = f(c) (the limit equals the function value).
If any one condition fails, f is discontinuous at c. Exam questions often ask you to state which condition fails.
Hole (removable): The limit exists but f(c) is either undefined or equals a different value. Conditions 1 or 3 fail.
Jump: Left-hand and right-hand limits exist but are not equal. Condition 2 fails.
Infinite (asymptotic): The function blows up to ±∞ near c. Condition 2 fails.
A rational function f(x) = p(x)/q(x) is continuous everywhere except where q(x) = 0.
Example: f(x) = (x + 4)/(x − 2) is continuous everywhere except at x = 2.
The standard exam question gives a piecewise function and asks either:
"Is f continuous at the boundary?" Check all three steps.
"Find the value of A that makes f continuous." Set the left-hand limit equal to the right-hand limit at the boundary and solve for A.
Example: f(x) = { x³ − x, if x < 2; Ax + 3, if x ≥ 2 }.
Step 1: f(2) = A(2) + 3 = 2A + 3.
Step 2: Left-hand limit = (2)³ − 2 = 6. Right-hand limit = 2A + 3.
Step 3: Set 6 = 2A + 3. Solve: 2A = 3, so A = 3/2.
Another example: f(x) = { x² − 3, if x < 1; Ax + 5, if x ≥ 1 }.
Left-hand limit = (1)² − 3 = −2.
Right-hand limit = A(1) + 5 = A + 5.
Set −2 = A + 5. Solve: A = −7.
For f(x) = (x − 3)/[(x + 1)(x − 9)], discontinuities occur at x = −1 and x = 9 (where the denominator is zero). The function is continuous at every other real number.
\lim_{x \to c} f(x) = L \quad \text{if and only if} \quad \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = LHorizontal asymptote rule for rational functions:
Given f(x) = (a_n x^n + ...)/(b_m x^m + ...):
If n < m, the horizontal asymptote is y = 0.
If n = m, the horizontal asymptote is y = a_n / b_m.
If n > m, there is no horizontal asymptote.
Rewriting expressions in ax^n form (for later differentiation):
3/x² = 3x⁻²
3x²√x = 3x^(5/2)
−5/(⁵√x⁷) = −5x^(−7/5)
√x / x² = x^(−3/2)
Difference quotient (slope of secant line):
\frac{f(x_2) - f(x_1)}{x_2 - x_1} \quad \text{or equivalently} \quad \frac{f(a+h) - f(a)}{h}This is the foundation for the derivative definition in the next unit.
"If f(c) is undefined, the limit does not exist." Wrong. A limit depends on the values near c, not at c. A hole in the graph means f(c) may be undefined, but the limit can still be a perfectly good number.
"The limit equals f(c)." Only when f is continuous at c. If there is a hole or jump, the limit and f(c) can differ (or f(c) might not exist at all).
"0/0 means the limit is 0 or undefined." 0/0 is an indeterminate form, which means you need to do more work. The actual limit could be any real number, or it could be ±∞, or it could not exist.
"A function cannot cross a horizontal asymptote." It can. A horizontal asymptote describes end behaviour (as x → ±∞). The function may cross that line at finite x-values.
Confusing a vertical asymptote with a hole. Both involve a zero in the denominator. If the same factor cancels from the numerator, it is a hole; if it does not cancel, it is a vertical asymptote.
⚠️ The three-step continuity check is a near-guaranteed exam question. Memorise the steps and practise stating which step fails.
⚠️ Piecewise functions with an unknown constant A: set left-hand limit = right-hand limit and solve. Quick points if you have the method down.
⚠️ Limits involving factoring or rationalising appear on every exam. If direct substitution gives 0/0, do not write "undefined"; simplify first.
⚠️ Horizontal asymptote questions test whether you can compare polynomial degrees quickly. Drill the three cases until they are automatic.
⚠️ Graphical limit reading: expect a graph with open/closed circles and be asked for one-sided limits, the two-sided limit, and f(c) separately.
True or false: If lim (x→5⁻) f(x) = 3 and lim (x→5⁺) f(x) = 7, then lim (x→5) f(x) = 5. False. The limit DNE because the one-sided limits disagree.
Fill in the blank: For a rational function where the degree of the numerator equals the degree of the denominator, the horizontal asymptote is y = ______. Ratio of leading coefficients.
True or false: 0/0 is a valid final answer for a limit. False. It is an indeterminate form requiring further simplification.
Fill in the blank: The three conditions for continuity at x = c are (1) f(c) is ______, (2) lim (x→c) f(x) ______, and (3) ______. Defined; exists; lim (x→c) f(x) = f(c).
True or false: A function can be continuous at a point where it has a vertical asymptote. False. A vertical asymptote means the function is unbounded and discontinuous there.
Q: Evaluate lim (x→2) (x² − 4)/(x − 2).
A: Factor the numerator: (x − 2)(x + 2)/(x − 2) = x + 2. Substituting x = 2 gives 4.
Q: Evaluate lim (x→∞) (2x² + 5x − 7)/(x − 4).
A: Degree of numerator (2) > degree of denominator (1), so the limit is ±∞. Since leading coefficients are positive, the limit is +∞. There is no horizontal asymptote.
Q: Find all vertical and horizontal asymptotes of f(x) = (x − 1)/(x² − 1).
A: Factor: (x − 1)/[(x − 1)(x + 1)] = 1/(x + 1) after cancellation. The cancelled factor gives a hole at x = 1 (not an asymptote). Vertical asymptote at x = −1. Horizontal asymptote: degree 0 over degree 1, so y = 0.
Q: Let f(x) = { x² − 3 if x < 1; Ax + 5 if x ≥ 1 }. Find A so that f is continuous.
A: Left-hand limit at x = 1: (1)² − 3 = −2. Right-hand limit: A(1) + 5 = A + 5. Set −2 = A + 5, so A = −7.
Q: From a graph, lim (x→2⁻) f(x) = 7 and lim (x→2⁺) f(x) = 13. What is lim (x→2) f(x)?
A: The two-sided limit does not exist (DNE) because the one-sided limits are not equal.
Q: Evaluate lim (x→8) (x² − 64)/(x − 8).
A: Factor: (x − 8)(x + 8)/(x − 8) = x + 8. Substituting x = 8 gives 16.
limit, one-sided limit, left-hand limit, right-hand limit, two-sided limit, limit does not exist, DNE, continuity, continuous function, discontinuity, removable discontinuity, jump discontinuity, infinite discontinuity, piecewise function, vertical asymptote, horizontal asymptote, indeterminate form, 0/0, factoring limits, rationalising limits, conjugate method, direct substitution, limit laws, limit at infinity, end behaviour, rational function asymptotes, three-step continuity test, finding A for continuity, MAC 2233, calculus for business, FIU, sections 2.1 2.2 2.3 2.4 2.5
Connections to other topics: Limits lead directly into the definition of the derivative (section 2.7 and beyond), which is defined as a limit of the difference quotient. Continuity matters because most differentiation rules assume the function is continuous. Horizontal asymptotes reappear when you analyse long-run behaviour of cost, revenue, and profit functions in business calculus applications.