Difficulty: Introductory to Intermediate | Prerequisites: Pre-calculus, familiarity with function notation and basic trigonometry.
Limits describe the value a function approaches as the input gets close to a particular point, even if the function is undefined there. Continuity means the function's limit and its actual value agree, so there is no break in the graph. The Intermediate Value Theorem guarantees that a continuous function on a closed interval hits every y-value between its endpoints, which is the standard tool for proving a solution exists inside a given interval.
Limit
The value that f(x) approaches as x approaches a given number c. Written lim(x→c) f(x) = L. In simple terms, this means: "Where is the function headed?" You are not asking what happens at x = c, only what happens as you get closer and closer.
One-sided limit
The value f(x) approaches from only one direction. lim(x→c⁻) f(x) is the left-hand limit; lim(x→c⁺) f(x) is the right-hand limit. Think of it as: approaching a doorway from only one side. The two-sided limit exists only when both one-sided limits agree.
Continuity at a point
A function f 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, this means: you can draw the graph through that point without lifting your pen.
Removable discontinuity
A point where the limit exists but either f(c) is undefined or f(c) does not equal the limit. The "hole" can be filled by redefining f(c) to equal the limit. Think of it as: a single missing dot on an otherwise smooth curve. You patch it by plugging in the right y-value.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any value between f(a) and f(b), then there exists at least one c in (a, b) with f(c) = N. In simple terms, this means: a continuous function cannot skip over a y-value. If f goes from 2 to 7, it must pass through every number between 2 and 7 at some point.
Cancellation method (for evaluating limits)
A technique for 0/0 indeterminate forms where you factor the numerator and denominator, cancel the common factor, and then substitute. Think of it as: the algebraic trick that removes the hole so you can substitute directly.
Direct substitution: If f is continuous at c, just plug in: lim(x→c) f(x) = f(c). Always try this first.
Cancellation (factoring): When substitution gives 0/0, factor and cancel. Example: lim(x→1) (x² – 1)/(x – 1) = lim(x→1) (x + 1)(x – 1)/(x – 1) = lim(x→1) (x + 1) = 2.
Special trigonometric limit: lim(u→0) sin(u)/u = 1. This is used heavily in derivative proofs and in exam problems. To evaluate lim(x→0) sin(5x)/(3x), rewrite as (5/3) · [sin(5x)/(5x)] and use the standard result to get 5/3.
Limit laws: Limits respect addition, subtraction, multiplication, division (provided the denominator's limit is nonzero), and constant multiples. These let you break complicated limits into simpler pieces.
A function is continuous on an interval if it is continuous at every point in that interval.
Polynomials, rational functions (where defined), trigonometric functions, exponentials, and logarithms are continuous on their domains.
If f has a removable discontinuity at x = c (the limit exists but f(c) is wrong or missing), you fix it by defining f(c) = lim(x→c) f(x). For f(x) = (x² – 1)/(x – 1) at x = 1, the limit is 2, so setting f(1) = 2 makes the function continuous.
Requirements: f must be continuous on the closed interval [a, b].
Conclusion: for any target value N between f(a) and f(b), there is some c in (a, b) with f(c) = N.
Typical exam use: you are given a table of values and asked which interval must contain a root or a particular y-value. Find an interval where f changes from below N to above N (or vice versa). Continuity plus that sign change guarantees a solution inside.
Example from the mock exam: f(0) = 10.1 and f(1) = 3.4, and the target is f(c) = 4. Because 4 lies between 3.4 and 10.1 and f is continuous, the IVT guarantees a c in (0, 1).
Defined as [f(b) – f(a)] / (b – a). This is the slope of the secant line between two points.
Example: sales go from 512 in 2010 to 2,612 in 2016. Average rate of change = (2612 – 512)/(2016 – 2010) = 2100/6 = 350 games per year.
\lim_{u \to 0} \frac{\sin u}{u} = 1\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}\text{Continuity at } c: \quad f(c) \text{ exists}, \quad \lim_{x \to c} f(x) \text{ exists}, \quad \lim_{x \to c} f(x) = f(c)\text{IVT: } f \text{ continuous on } [a,b],\; N \text{ between } f(a) \text{ and } f(b) \;\Rightarrow\; \exists\, c \in (a,b) \text{ with } f(c) = NStudents often think the limit of f at c equals f(c). It does not, unless f is continuous there. A function can have a limit of 2 at x = 1 while f(1) = 7 or f(1) is undefined.
Students confuse "the limit does not exist" with "the function is undefined." These are different things. A function can be undefined at c while the limit at c exists perfectly well (that is exactly what a removable discontinuity is).
When using the IVT, students sometimes forget to verify continuity. The theorem requires f to be continuous on the closed interval. Without continuity, a function can jump right over the target value.
Students sometimes try to "cancel" terms that are added rather than multiplied. You can only cancel a common factor from the numerator and denominator, not a common summand.
⚠️ The special limit lim(u→0) sin(u)/u = 1 appears directly in multiple-choice questions. Know how to adapt it when the argument is scaled (e.g., sin(5x)/(3x)).
⚠️ The cancellation method for 0/0 limits (factor, cancel, substitute) is tested in both multiple-choice and short-answer sections.
⚠️ IVT questions give you a table of function values and ask you to identify which interval contains a root or a particular y-value. The key step is checking that the target value lies between f(a) and f(b).
⚠️ Removable discontinuity questions ask what value to assign to f(c) to make f continuous. The answer is always the limit. This appeared as both a multiple-choice concept question and a short-answer computation.
⚠️ Average rate of change is a secant-line slope calculation. Watch your units.
True or false: If lim(x→3) f(x) = 5, then f(3) must equal 5. (False. The limit can differ from the function value.)
Fill in the blank: lim(u→0) sin(u)/u = ___. (1)
True or false: The IVT can be applied to any function, continuous or not. (False. Continuity on the closed interval is required.)
Fill in the blank: To remove a removable discontinuity at x = c, define f(c) = ___. (The limit of f as x approaches c.)
True or false: lim(x→1) (x² – 1)/(x – 1) does not exist because the function is undefined at x = 1. (False. The limit is 2.)
Q: Evaluate lim(x→0) sin(5x)/(3x).
A: Rewrite as (5/3) · [sin(5x)/(5x)]. As x→0, sin(5x)/(5x)→1, so the limit is 5/3.
Q: A continuous function has f(0) = 10.1 and f(1) = 3.4. Which interval is guaranteed by the IVT to contain a c with f(c) = 4?
A: The interval (0, 1), because 4 lies between 3.4 and 10.1 and f is continuous.
Q: Evaluate lim(x→1) (x² – 1)/(x – 1) using the cancellation method.
A: Factor: (x² – 1) = (x + 1)(x – 1). Cancel (x – 1). Substitute x = 1 into (x + 1) to get 2.
Q: Define f(x) = (x² – 1)/(x – 1) for x ≠ 1 and set f(1) = 2. Is f continuous at x = 1?
A: Yes. The limit as x→1 is 2 (by cancellation), f(1) is defined as 2, and the limit equals the defined value, so all three conditions for continuity hold.
Q: What value must be assigned to f(1) to make f(x) = (x² – 1)/(x – 1) continuous at x = 1?
A: f(1) = 2, because that is the limit of f as x approaches 1.
Q: A shop sold 512 games in 2010 and 2,612 games in 2016. What is the average rate of change in annual sales?
A: (2612 – 512)/(2016 – 2010) = 2100/6 = 350 games per year.
Limits are the foundation of the derivative. The definition of the derivative is itself a limit (lim(h→0) [f(x+h) – f(x)]/h), so everything you learn here feeds directly into differentiation. The IVT connects to optimisation: the Extreme Value Theorem (continuous functions on closed intervals have absolute maxima and minima) relies on the same continuity requirement. Average rate of change is the precursor to instantaneous rate of change, which is the derivative.
limit, one-sided limit, left-hand limit, right-hand limit, limit laws, continuity, continuous function, removable discontinuity, jump discontinuity, infinite discontinuity, hole in a graph, Intermediate Value Theorem, IVT, existence of roots, sign change, cancellation method, factoring limits, 0/0 indeterminate form, sin(u)/u, special trig limit, average rate of change, secant line slope, MTH 132, Calculus I, Michigan State University