Difficulty: Introductory to Intermediate | Prerequisites: Algebra, basic trigonometry, familiarity with functions and their graphs.
Limits and continuity form the foundation of everything in Calculus I. Before you can differentiate or integrate, you need to understand what it means for a function to approach a value and what it means for it to behave predictably at a point. This topic sits right at the start of the course, but its ideas underpin every theorem you will meet later, from the definition of the derivative to the Fundamental Theorem of Calculus. If you skipped precalculus or feel shaky on factoring and rationalising, revisit those first.
A limit describes what value a function approaches as the input gets close to some number, even if the function is undefined there. Continuity means no breaks, holes, or jumps in the graph. Mastering limit evaluation techniques (factoring, rationalising, L'Hopital's Rule) and knowing the three conditions for continuity will carry you through most of the exam questions on this topic.
Limit
The value that f(x) approaches as x approaches a particular number c. Written lim(x->c) f(x) = L. In simple terms, it answers: "What is this function heading towards, even if it never gets there?"
One-sided limit
The value f(x) approaches from only the left (x -> c-) or only the right (x -> c+). A two-sided limit exists only when both one-sided limits agree. Think of it as approaching a point from one direction only, like walking toward a door from the hallway versus from the room.
Continuity
A function f(x) is continuous at x = c when three conditions hold: f(c) is defined, lim(x->c) f(x) exists, and lim(x->c) f(x) = f(c). In simple terms, you can draw the function through that point without lifting your pen.
Removable discontinuity (hole)
A point where the limit exists but either f(c) is undefined or f(c) differs from the limit. The graph has a "hole" you could fill by redefining f(c). Think of it as a single missing dot on an otherwise smooth curve.
Infinite discontinuity (vertical asymptote)
A point where f(x) grows without bound (toward positive or negative infinity) as x approaches c. The graph shoots up or down and never crosses x = c. This is the behaviour you see at x = 0 for f(x) = 1/x.
Jump discontinuity
A point where the left-hand and right-hand limits both exist but are not equal. The graph "jumps" from one value to another. Piecewise functions are the typical source of these.
Indeterminate form
An expression like 0/0 or infinity/infinity that does not have a value on its own. It signals that algebraic simplification or L'Hopital's Rule is needed to find the limit. In simple terms, the function is doing something ambiguous at that point and you need another technique to resolve it.
L'Hopital's Rule
If lim(x->c) f(x)/g(x) produces an indeterminate form 0/0 or infinity/infinity, the limit equals lim(x->c) f'(x)/g'(x), provided this new limit exists. Think of it as a shortcut: instead of algebraic tricks, differentiate top and bottom separately and try the limit again.
Vertical asymptote
A vertical line x = c where f(x) tends to positive or negative infinity. Formally, lim(x->c) f(x) = infinity (or negative infinity). This is where the denominator of a rational function hits zero and the numerator does not.
When direct substitution gives 0/0, factor numerator and denominator, cancel the common factor, then substitute again.
Example: f(x) = (x^2 - 9)/(x - 3). Factor the numerator as (x - 3)(x + 3). Cancel (x - 3) to get x + 3. Then lim(x->3) (x + 3) = 6.
The function is undefined at x = 3, but the limit still equals 6 because we are asking what the function approaches, not what it equals there.
When the expression involves a square root and direct substitution gives 0/0, multiply numerator and denominator by the conjugate.
Example: lim(x->4) (sqrt(x) - 2)/(x - 4). Multiply top and bottom by (sqrt(x) + 2). The numerator becomes x - 4, which cancels with the denominator, leaving 1/(sqrt(x) + 2). Substituting x = 4 gives 1/4.
Compare the highest powers (degrees) of x in the numerator and denominator.
Same degree: the limit is the ratio of the leading coefficients. Example: lim(x->infinity) (4x^3 - 2x + 1)/(2x^3 + 5x^2) = 4/2 = 2.
Numerator degree higher: the limit is infinity (or negative infinity).
Denominator degree higher: the limit is 0.
lim(x->0) sin(x)/x = 1. This is a foundational result used throughout calculus.
lim(x->0) (e^x - 1)/x = 1. This follows from L'Hopital's Rule: differentiate top and bottom to get e^x/1, then substitute x = 0.
Applies only when direct substitution produces an indeterminate form (0/0 or infinity/infinity).
Differentiate the numerator and denominator separately (this is not the quotient rule).
Re-evaluate the limit. If it is still indeterminate, apply L'Hopital's Rule again.
Example: lim(x->0) (e^x - 1)/x gives 0/0. Differentiating: e^x/1. At x = 0, this is 1.
Removable (hole): the limit exists, but f(c) is either undefined or disagrees with the limit. Example: f(x) = (x^2 - 9)/(x - 3) at x = 3.
Infinite (vertical asymptote): f(x) blows up to infinity or negative infinity. Example: f(x) = 1/(x - 2) at x = 2.
Jump: the left and right limits exist but differ. Example: a piecewise function that equals 2x + 3 for x <= 1 and kx - 1 for x > 1, if k is chosen so the pieces do not meet.
A function f(x) has a vertical asymptote at x = c when lim(x->c) f(x) = infinity or negative infinity.
To find vertical asymptotes of a rational function, set the denominator equal to zero and confirm the numerator is nonzero there.
f(c) is defined.
lim(x->c) f(x) exists.
lim(x->c) f(x) = f(c).
All three must hold. If any one fails, f is discontinuous at c.
For a piecewise function to be continuous at the boundary x = c, the value from the left piece must equal the value from the right piece, and both must equal f(c).
Example: f(x) = {2x + 3 if x <= 1, kx - 1 if x > 1}. For continuity at x = 1, set 2(1) + 3 = k(1) - 1. So 5 = k - 1, giving k = 6.
lim(x->0) sin(x)/x = 1
lim(x->0) (e^x - 1)/x = 1
L'Hopital's Rule: if lim f(x)/g(x) is 0/0 or infinity/infinity, then lim f(x)/g(x) = lim f'(x)/g'(x)
For rational functions as x -> infinity: compare leading-term degrees. Same degree -> ratio of leading coefficients. Higher degree on top -> infinity. Higher degree on bottom -> 0.
Three conditions for continuity at c: (1) f(c) defined, (2) lim(x->c) f(x) exists, (3) lim(x->c) f(x) = f(c).
Limits describe any situation where you care about a trend rather than an exact value: the speed a falling object is approaching at a given instant, the temperature a cooling cup of tea is heading towards, or the concentration a chemical reaction is settling at over time. Engineers use limits when designing control systems that need to converge on a target value without overshooting.
Students often think that if a function is undefined at x = c, the limit at c cannot exist. The limit describes approach, not arrival. The function f(x) = (x^2 - 9)/(x - 3) is undefined at x = 3, but lim(x->3) f(x) = 6.
Students confuse the limit existing with the function being continuous. The limit can exist at a point where the function has a hole (removable discontinuity). Continuity requires the extra step of the limit equalling f(c).
Students apply L'Hopital's Rule when the form is not indeterminate. If direct substitution gives a finite, non-zero value, L'Hopital's Rule does not apply and will give the wrong answer.
Students sometimes treat L'Hopital's Rule as if it were the quotient rule. You differentiate the numerator and denominator independently, not as a single fraction.
Expect at least one question asking you to classify a discontinuity (removable, infinite, or jump). Know the definitions cold.
Factoring limits like (x^2 - 9)/(x - 3) is an exam staple. Practise until you can do it on autopilot.
The three conditions for continuity are frequently tested as a "state the conditions" short-answer question.
L'Hopital's Rule questions almost always use e^x or ln(x). Recognise the 0/0 or infinity/infinity form before applying.
Piecewise continuity problems ("find k such that f is continuous") appear regularly. Set the two pieces equal at the boundary and solve.
lim(x->0) sin(x)/x = 1 is a result you must know by heart. It appears directly or as a building block in other limit problems.
True or False: If f(3) is undefined, then lim(x->3) f(x) does not exist. (False.)
Fill in the blank: lim(x->0) sin(x)/x = ___. (1.)
True or False: A removable discontinuity means the left and right limits differ. (False. It means the limit exists but does not equal f(c), or f(c) is undefined.)
Fill in the blank: For continuity at x = c, three conditions are: f(c) is defined, lim(x->c) f(x) ___, and lim(x->c) f(x) = ___. (exists; f(c).)
True or False: L'Hopital's Rule can be used whenever you have a fraction. (False. Only for indeterminate forms 0/0 or infinity/infinity.)
Q: Consider f(x) = (x^2 - 9)/(x - 3) for x != 3. What is lim(x->3) f(x)?
A: Factor the numerator: (x - 3)(x + 3)/(x - 3) = x + 3. Substituting x = 3 gives 6.
Q: What type of discontinuity exists at x = 3 for f(x) = (x^2 - 9)/(x - 3)?
A: Removable discontinuity (hole). The limit exists (equals 6) but f(3) is undefined.
Q: Evaluate lim(x->infinity) (4x^3 - 2x + 1)/(2x^3 + 5x^2).
A: Same degree (3) in numerator and denominator. Ratio of leading coefficients: 4/2 = 2.
Q: Find lim(x->0) sin(x)/x.
A: 1. This is a standard limit.
Q: Using L'Hopital's Rule, find lim(x->0) (e^x - 1)/x.
A: Direct substitution gives 0/0. Differentiate: e^x / 1. At x = 0, this equals 1.
Q: Evaluate lim(x->4) (sqrt(x) - 2)/(x - 4) using algebraic simplification.
A: Multiply by the conjugate (sqrt(x) + 2)/(sqrt(x) + 2). Numerator becomes x - 4. Cancel with denominator: 1/(sqrt(x) + 2). At x = 4: 1/(2 + 2) = 1/4.
Q: A function f(x) has a vertical asymptote at x = c. Which condition holds?
A: lim(x->c) f(x) = infinity (or negative infinity).
Q: State the three conditions for f(x) to be continuous at x = c.
A: (1) f(c) is defined. (2) lim(x->c) f(x) exists. (3) lim(x->c) f(x) = f(c).
Q: Find k such that f(x) = {2x + 3 for x <= 1, kx - 1 for x > 1} is continuous at x = 1.
A: Left value at x = 1: 2(1) + 3 = 5. Right limit at x = 1: k(1) - 1 = k - 1. Set equal: k - 1 = 5, so k = 6.
Limits are the machinery behind derivatives: the derivative is defined as lim(h->0) [f(x+h) - f(x)]/h, so every differentiation rule you learn later is, at bottom, a limit result. Continuity matters again when you reach the Intermediate Value Theorem and the Extreme Value Theorem, both of which require the function to be continuous on a closed interval. The Fundamental Theorem of Calculus also relies on continuity of the integrand.
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