Difficulty: Introductory | Prerequisites: College algebra, function notation, factoring, rationalising expressions.
Limits are the gateway concept of calculus. They describe what value a function approaches as its input gets close to some number, even if the function never reaches that value. Everything that follows in the course (derivatives, integrals, optimisation) is built on limits. Continuity, which depends on limits, tells you whether a function has any breaks, jumps, or holes, and determines whether familiar rules (like "just plug in") will work. If you are catching up, make sure you are comfortable evaluating functions, factoring quadratics, and rationalising denominators before diving in.
A limit asks: "What y-value does f(x) approach as x gets close to c?" You find it by plugging in directly, factoring, or using conjugates. If the left and right limits disagree, the limit does not exist. Continuity at a point requires the function to be defined there, the limit to exist, and the two to match.
Limit
The y-value that f(x) approaches as x gets closer and closer to a particular number c. The function does not need to equal that value at c; it only matters what happens nearby.
In simple terms: "Where is the graph heading as you slide along towards c?"
Left-hand limit (lim x→c⁻)
The value f(x) approaches when x comes from numbers less than c (from the left on the number line).
Think of it as: approaching c using values like 1.9, 1.99, 1.999 when c = 2.
Right-hand limit (lim x→c⁺)
The value f(x) approaches when x comes from numbers greater than c (from the right on the number line).
Think of it as: approaching c using values like 2.1, 2.01, 2.001 when c = 2.
Does Not Exist (DNE)
A limit does not exist when the left-hand and right-hand limits are not equal to each other, or when the function oscillates without settling. Also used when a function value is undefined at a point.
Indeterminate form (0/0)
The result you get when direct substitution into a rational expression gives 0 in both the numerator and the denominator. This does not mean the limit is zero or undefined; it means more work is needed (factoring or conjugate multiplication) to find the true limit.
In simple terms: 0/0 is a signal that the expression can be simplified further.
Vertical asymptote
A vertical line x = c where the function grows without bound (towards ∞ or −∞) as x approaches c. This happens when plugging in gives the form n/0, where n ≠ 0.
Think of it as: the graph shoots up or down towards a wall it can never cross.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x goes to ∞ or −∞. Found by evaluating the limit of the function at infinity.
Think of it as: the long-run behaviour of the graph, far out to the right or left.
Hole (removable discontinuity)
A single missing point on the graph where the function is undefined, but the limit still exists. Arises from the 0/0 indeterminate form, after cancelling the common factor. The y-coordinate of the hole is the value of the simplified function at that x.
Reciprocal limit
The important result that lim (k / xᵖ) = 0 as x → ±∞ for any constant k and positive power p. Used constantly when evaluating rational functions at infinity.
Continuity at a point
A function f(x) is continuous at x = c when three conditions hold: (I) f(c) is defined, (II) the limit as x→c exists, and (III) the limit equals f(c). Informally, you can draw through the point without lifting your pen.
In simple terms: no holes, no jumps, no asymptotes at that point.
A limit describes the y-value a function approaches, not necessarily the value it equals. You read lim x→c f(x) = L as "the limit of f(x) as x approaches c is L."
Left-hand limit: lim x→c⁻ f(x) considers values of x less than c (approaching from the left).
Right-hand limit: lim x→c⁺ f(x) considers values of x greater than c (approaching from the right).
Two-sided limit: lim x→c f(x) exists if and only if the left-hand limit equals the right-hand limit. If they disagree, the limit DNE.
Reading limits from a graph:
To find lim x→c⁻ f(x), trace the curve from the left side towards x = c and read the y-value the curve is heading towards.
To find lim x→c⁺ f(x), trace from the right.
The actual value f(c), shown by a filled dot, may differ from the limit. An open circle means the function is not defined at that point.
A limit can exist even when f(c) is undefined. A limit can also fail to exist even when f(c) is perfectly well defined.
Notation rule from your instructor: never write a limit symbol without a function or expression after it. Writing "lim x→c = 3" is incorrect. Always write "lim x→c f(x) = 3."
When c is a finite real number, the first move is always to try plugging x = c directly into f(x).
Polynomials: Direct substitution always works because polynomials are continuous everywhere.
Example: lim x→3 (x + 5) = 3 + 5 = 8.
Rational functions: Direct substitution works provided the denominator does not become zero at x = c.
Example: lim x→9 (√x − 3)/(x + 1) = (√9 − 3)/(9 + 1) = 0/10 = 0.
Constants: The limit of a constant is just the constant itself. lim x→c k = k for any real number k.
Example: lim x→8 72 = 72.
Piecewise functions:
When f(x) is defined by different rules on different intervals, use the rule that applies to the direction you are approaching from.
For lim x→c⁻, use the formula for values less than c.
For lim x→c⁺, use the formula for values greater than c.
For f(c), use whichever rule explicitly includes x = c.
The two-sided limit exists only if the left and right limits agree.
Example: given f(x) = { x² + x if x < 2, 7 if x = 2, 3 − x if x > 2 }.
lim x→2⁻ f(x) = 2² + 2 = 6 (using x² + x)
lim x→2⁺ f(x) = 3 − 2 = 1 (using 3 − x)
Because 6 ≠ 1, lim x→2 f(x) = DNE.
f(2) = 7 (using the middle rule).
For a value away from the boundary, use whichever single rule covers it. For instance, lim x→−1 f(x) = (−1)² + (−1) = 0 using the first rule, since −1 < 2.
When direct substitution gives 0/0, the limit is not zero and is not undefined. It means the numerator and denominator share a common factor that can be cancelled. After simplifying, substitute again.
Method 1: Factoring
Factor numerator and denominator, cancel the shared factor, then substitute.
Example: lim x→1 (x² − 1)/(x − 1).
Direct substitution: (1 − 1)/(1 − 1) = 0/0 (indeterminate).
Factor: (x − 1)(x + 1)/(x − 1) = x + 1 after cancelling.
Now substitute: 1 + 1 = 2.
The limit is 2.
Method 2: Conjugate multiplication
Used when the expression contains a square root. Multiply numerator and denominator by the conjugate of whichever part contains the root.
Recall: the conjugate of (a + b) is (a − b), and vice versa.
Example: lim x→64 (√x − 8)/(x − 64).
Direct substitution: (√64 − 8)/(64 − 64) = 0/0.
Multiply top and bottom by (√x + 8): (x − 64)/[(x − 64)(√x + 8)] = 1/(√x + 8).
Substitute x = 64: 1/(√64 + 8) = 1/16.
What 0/0 tells you graphically:
The function has a hole at x = c. The function is undefined there, but the graph behaves as though the point exists. The y-coordinate of the hole is the value of the limit.
When direct substitution gives a non-zero number divided by zero (n/0 where n ≠ 0), the function blows up. The limit will be ∞, −∞, or DNE, and the graph has a vertical asymptote at x = c.
You must use a table to check what happens from the left and right separately.
Three possible outcomes:
Both sides go to +∞: lim x→c f(x) = ∞.
Both sides go to −∞: lim x→c f(x) = −∞.
One side goes to +∞ and the other to −∞: lim x→c f(x) = DNE.
Example: lim x→5 (x + 1)/(x − 5).
Direct substitution gives 6/0, so this is n/0.
Build a table with values near 5 from each side:
x = 4.9 → 5.9/(−0.1) = −59 (large negative)
x = 4.99 → 5.99/(−0.01) = −599
x = 5.01 → 6.01/0.01 = 601
x = 5.001 → 6.001/0.001 = 6001
From the left the function goes to −∞. From the right it goes to +∞. Since these disagree, lim x→5 (x + 1)/(x − 5) = DNE, and x = 5 is a vertical asymptote.
Example: lim x→−3 −1/(x + 3)².
Direct substitution gives −1/0, so n/0.
Because (x + 3)² is always positive near x = −3, the fraction is always negative. Both sides approach −∞. So the limit is −∞.
Polynomial functions: The end behaviour of a polynomial is determined entirely by its leading term.
lim x→±∞ f(x) = lim x→±∞ (leading term)
Example: lim x→−∞ (−2x⁵ − 4x³ + 2) = lim x→−∞ (−2x⁵). Since the degree is odd and the coefficient is negative, as x → −∞ the leading term goes to +∞. So the limit is +∞.
Example: lim x→∞ (3x⁴ + 2x² − 10) = lim x→∞ 3x⁴ = +∞ (even degree, positive coefficient).
Rational functions: Compare the leading terms of the numerator and denominator.
lim x→±∞ p(x)/q(x) = lim x→±∞ (leading term of top)/(leading term of bottom)
Simplify and cancel common powers of x. Three cases arise:
Degree of top < degree of bottom: x’s remain in the denominator only. The limit equals 0. This is the reciprocal limit: lim k/xᵖ = 0.
Example: lim x→+∞ (x⁵ + 3x − 1)/(x⁹ − 4x + 5) = lim x⁵/x⁹ = lim 1/x⁴ = 0.
Degree of top = degree of bottom: All x’s cancel, leaving a ratio of the leading coefficients.
Example: lim x→−∞ (−2x⁴ − 3x + 1)/(x² − 5)² = lim −2x⁴/x⁴ = −2.
Note: (x² − 5)² expands to have leading term x⁴.
Degree of top > degree of bottom: x’s remain in the numerator. Treat it like a polynomial limit: the answer is ∞ or −∞.
Example: lim x→−∞ (5x¹¹ − 2x⁴ + 7x − 2)/(x⁷ + 6x − 2) = lim 5x¹¹/x⁷ = lim 5x⁴. As x → −∞, 5x⁴ → +∞ (even power).
Horizontal asymptote: Evaluate lim x→±∞ f(x). If the limit equals a finite number L, then y = L is a horizontal asymptote.
Vertical asymptote: Set the denominator equal to zero to find candidate x-values. For each candidate c, evaluate lim x→c f(x).
If plugging in gives 0/0: the common factor cancels, leaving a hole. The y-coordinate of the hole is the value of the simplified expression at x = c.
If plugging in gives n/0 (n ≠ 0): x = c is a vertical asymptote.
Worked example: Find all asymptotes and holes of f(x) = (x² − 2x − 3)/(x² − 1).
Factor: (x − 3)(x + 1) / [(x − 1)(x + 1)].
Denominator is zero at x = 1 and x = −1.
At x = −1: both numerator and denominator are zero (0/0). Cancel (x + 1) to get (x − 3)/(x − 1). Substitute x = −1: (−4)/(−2) = 2. There is a hole at (−1, 2).
At x = 1: numerator is (1 − 3)(1 + 1) = −4, denominator is 0. This is n/0, so x = 1 is a vertical asymptote.
Horizontal asymptote: lim x→±∞ x²/x² = 1. So y = 1 is the horizontal asymptote.
This type of problem (finding all asymptotes and holes using limits) appears on exams regularly. Your instructor requires limit notation for full credit.
A function f(x) is continuous at x = c if all three conditions hold:
Condition I: f(c) is defined (the function has a value at c).
Condition II: lim x→c f(x) exists (the left and right limits agree).
Condition III: lim x→c f(x) = f(c) (the limit and the function value match).
If any one of these fails, f is discontinuous at c. When explaining why a function is not continuous, state which condition fails using the actual definition (for example, "f(−4) is undefined" or "lim x→5 f(x) does not exist"), not just the condition number.
Proving continuity: To show f(x) = (x − 2)/(x + 1) is continuous at x = 1, check each condition.
I: f(1) = (1 − 2)/(1 + 1) = −1/2. Defined.
II: lim x→1 (x − 2)/(x + 1) = −1/2 (direct substitution works, denominator ≠ 0). Exists.
III: lim x→1 f(x) = −1/2 = f(1). They match.
All three conditions are met, so f is continuous at x = 1.
Finding A for piecewise continuity:
Given f(x) = { x² − 3 if x < 1, Ax + 5 if x ≥ 1 }, find A so that f is continuous everywhere.
The only place the formula changes is x = 1, so check continuity there.
I: f(1) = A(1) + 5 = A + 5 (use the x ≥ 1 rule). Defined for any A.
II: Left limit: lim x→1⁻ (x² − 3) = 1 − 3 = −2. Right limit: lim x→1⁺ (Ax + 5) = A + 5. For the limit to exist, set −2 = A + 5, giving A = −7.
III: With A = −7, f(1) = −7 + 5 = −2 = limit. Condition satisfied.
So A = −7 makes f continuous everywhere.
Important instructor notes:
Always state the final two-sided limit (not just the one-sided limits). The whole reason for computing left and right limits is to determine this.
Always end by stating whether the function is continuous or not at the given point. The whole reason for checking the three conditions is to answer this question.
Limit existence: lim x→c f(x) exists if and only if lim x→c⁻ f(x) = lim x→c⁺ f(x).
Direct substitution outcomes:
Result is a real number: that number is the limit.
Result is 0/0: indeterminate form. Simplify by factoring or conjugate, then substitute again.
Result is n/0 (n ≠ 0): vertical asymptote. Use a table to determine ∞, −∞, or DNE.
Polynomial end behaviour: lim x→±∞ f(x) = lim x→±∞ (leading term).
Rational function end behaviour: lim x→±∞ p(x)/q(x) = lim x→±∞ (leading term of p)/(leading term of q).
Reciprocal limit: lim x→±∞ k/xᵖ = 0 for any constant k and positive power p.
Continuity at x = c: (I) f(c) defined, (II) lim x→c f(x) exists, (III) lim x→c f(x) = f(c).
Horizontal asymptote: y = L where L = lim x→±∞ f(x).
Vertical asymptote: x = c where lim x→c f(x) = ±∞ (n/0 form).
Hole: occurs at x = c when 0/0 form resolves. Coordinates: (c, limit value after simplification).
Limits describe what happens at boundaries: engineers use them to model how temperature or pressure approaches equilibrium. In business, the horizontal asymptote of a cost function tells you the long-run average cost per unit as production scales up. Continuity matters whenever a model must not have sudden jumps, for instance in pricing curves or supply-demand models.
Students often think the limit of f(x) at c must equal f(c). It does not. The limit describes the trend, not the value.
Students assume 0/0 means the limit is zero or undefined. It means the expression can be simplified. The limit usually exists and is a finite number.
Students confuse "undefined" and "does not exist." A function value can be undefined (no output at that input). A limit does not exist when the left and right limits disagree.
Students forget to check both sides when the result is n/0. The limit might be ∞, −∞, or DNE depending on the signs from the left and right.
⚠️ The three-step continuity check (defined, limit exists, they match) will appear on every exam. Always conclude by stating whether the function is continuous or not.
⚠️ Finding asymptotes and holes using limits is a staple exam problem. Your instructor insists on limit notation for full credit.
⚠️ Piecewise continuity problems (finding A or another constant) are tested regularly.
⚠️ Never write a bare limit symbol. Always include the function: lim x→c f(x), not lim x→c.
True or false: If f(3) = 5, then lim x→3 f(x) = 5.
Fill in the blank: When direct substitution gives 0/0, this is called the ________ form.
True or false: A function can have a limit at a point where it is not defined.
Fill in the blank: If the degree of the numerator is less than the degree of the denominator, the limit at infinity is ________.
True or false: A vertical asymptote occurs wherever the denominator is zero.
Answers: 1. False (f(c) and the limit can differ). 2. Indeterminate. 3. True. 4. Zero. 5. False (it could be a hole if the numerator is also zero there).
Q: Evaluate lim x→5 (x − 5)/(x² − 25).
A: Factor the denominator: (x − 5)(x + 5). Cancel (x − 5) to get 1/(x + 5). Substitute x = 5: 1/10.
Q: For f(x) = (x² − 2x − 3)/(x² − 1), identify all holes and vertical asymptotes.
A: Factor: (x − 3)(x + 1)/[(x − 1)(x + 1)]. Hole at x = −1 (0/0 cancels; y-coordinate = (−1 − 3)/(−1 − 1) = 2). Vertical asymptote at x = 1 (gives n/0 after cancelling).
Q: Evaluate lim x→∞ (−x⁵ + 1).
A: Leading term is −x⁵. As x → ∞, −x⁵ → −∞. The limit is −∞.
Q: Evaluate lim x→4 (x − 4)/(√x − 2).
A: Direct substitution gives 0/0. Multiply top and bottom by (√x + 2): [(x − 4)(√x + 2)]/[(x − 4)] = √x + 2. Substitute x = 4: √4 + 2 = 4.
Q: For f(x) = { x² − 3 if x < 1, Ax + 5 if x ≥ 1 }, find A so that f is continuous everywhere.
A: Set left limit = right limit at x = 1. Left: 1 − 3 = −2. Right: A + 5. So A + 5 = −2, giving A = −7.
Limits feed directly into the definition of the derivative (Section 2.4), which is itself a limit: f′(x) = lim h→0 [f(x + h) − f(x)]/h. Continuity is a prerequisite for differentiability: if a function is not continuous at a point, it cannot have a derivative there. Horizontal asymptotes reappear in curve sketching (Sections 4.1, 4.4) when you analyse long-run behaviour.
limits, left-hand limit, right-hand limit, one-sided limit, two-sided limit, DNE, does not exist, indeterminate form, 0/0, n/0, direct substitution, factoring limits, conjugate method, rationalising, vertical asymptote, horizontal asymptote, hole, removable discontinuity, continuity, continuous function, piecewise function, reciprocal limit, end behaviour, polynomial limit, rational function limit, limit at infinity, limit to infinity, MAC 2233, business calculus, FIU, Section 2.1, Section 2.2, Section 2.3