Source: MTH 132, Michigan State University, Sections 1.8/2.1
Tags: continuity, discontinuity, removable discontinuity, jump discontinuity, infinite discontinuity, piecewise functions, continuous functions, MTH 132, Calculus I, MSU
Difficulty: Introductory to Intermediate | Prerequisites: Limits (Sec. 1.4 -- 1.6), basic trigonometry
Continuity is one of the foundational ideas in calculus, bridging the work you have done on limits to the definition of the derivative. A function is continuous at a point when its limit equals its actual value there, with no gaps, jumps or blowups. This topic matters because the Intermediate Value Theorem (which relies on continuity) is a standard exam tool, and because you will need continuity when you move on to differentiation. If you are comfortable evaluating limits and know the basic trig values on [0, 2pi], you are ready.
A function is continuous at a point when you can draw through it without lifting your pencil: the limit exists, the function is defined and they match. Discontinuities come in three flavours (removable, jump, infinite), and you can force continuity for a piecewise function by choosing parameters that make the pieces agree at the boundary.
Continuous at a point
A function f is continuous at x = a when three conditions hold: f(a) is defined, lim(x->a) f(x) exists, and lim(x->a) f(x) = f(a). In simple terms, the graph has no hole, jump or vertical asymptote at that point.
Removable discontinuity
A point where the limit of f exists but either f is not defined there or f(a) does not equal the limit. Think of it as a single hole in the graph: if you "filled in" the missing dot, the function would be continuous.
Jump discontinuity
A point where the left-hand limit and the right-hand limit both exist but are not equal: lim(x->a-) f(x) ≠ lim(x->a+) f(x). In simple terms, the graph makes a sudden step, like a staircase.
Infinite discontinuity
A point where the function heads to +∞ or -∞ (a vertical asymptote). The limit does not exist because the output grows without bound.
Piecewise function
A function defined by different formulas on different intervals. To check continuity, you check whether the formulas agree (give the same value) at each boundary point.
Continuous on an interval
f is continuous on [a, b] when it is continuous at every interior point, continuous from the right at a, and continuous from the left at b. The interval is the largest connected set on which the function has no discontinuities.
Start by identifying where the function could break: denominators equal to zero, square roots of negative numbers, domain restrictions.
For f(x) = 1 / (1 + sin x) on [0, 2pi]:
Set the denominator to zero: 1 + sin x = 0, so sin x = -1, giving x = 3pi/2.
The function is continuous everywhere on [0, 2pi] except at x = 3pi/2.
Continuous on [0, 3pi/2) ∪ (3pi/2, 2pi].
Strategy: always ask "what makes the denominator zero?" or "what leaves the expression undefined?" first, then remove those points from the interval.
Removable discontinuity
The two-sided limit exists at x = a, but f(a) is either undefined or does not match the limit.
Graphically: a single hole (open circle) in the curve, possibly with a filled dot at a different height.
Example: f(x) = (x^2 - 1)/(x - 1) at x = 1. The limit is 2, but f(1) is undefined.
Jump discontinuity
The left-hand and right-hand limits both exist but disagree: lim(x->a-) f(x) ≠ lim(x->a+) f(x).
The two-sided limit therefore does not exist.
Graphically: the curve arrives at one height from the left and a different height from the right.
Infinite discontinuity
The function blows up to ±∞ near x = a (vertical asymptote).
Neither one-sided limit is finite, so the two-sided limit does not exist.
Graphically: the curve races off toward positive or negative infinity on at least one side.
Given a piecewise function with an unknown parameter (e.g. c), set the two pieces equal at the boundary and solve.
Worked example: f(x) = { x^2 - 10 for x ≤ c ; 8x - 26 for x > c }.
At x = c the two pieces must agree: c^2 - 10 = 8c - 26.
Rearrange: c^2 - 8c + 16 = 0, which factors as (c - 4)^2 = 0.
So c = 4.
The key step is always: evaluate the left-hand piece and the right-hand piece at the boundary, set them equal, solve for the parameter.
When the boundary itself is unknown (fill-in-the-blank style), you may get more than one valid (a, b) pair. In the recitation example with blanks in f(x) = { x^2 - 17 for x ≤ a ; bx - 29 for x > a }, three solutions exist: (a, b) = (4, 7), (2, 8), (6, 8).
\text{Continuity at } x = a:\quad \lim_{x \to a} f(x) = f(a)Three sub-conditions packed into that one line:
f(a) is defined.
lim(x->a) f(x) exists (left-hand and right-hand limits agree).
The limit equals the function value.
For a piecewise function with boundary at x = c:
\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)In practice, this means: plug c into both formulas and set them equal.
Continuity is why engineers can trust that a bridge beam, loaded gradually, will not suddenly snap to a different deflection. If the load-deflection relationship were discontinuous, small changes in weight could produce wild, unpredictable movements. In signal processing, a discontinuous signal creates sharp edges that show up as high-frequency noise, which is why filters are designed to keep signals smooth (continuous).
Students often think "the limit exists" and "the function is continuous" mean the same thing. They do not. A removable discontinuity has an existing limit but is still not continuous because f(a) is missing or wrong.
Students confuse "undefined" with "discontinuous." A function is discontinuous at a point in its domain (or at an endpoint of a stated interval). If the point is simply not in the domain at all, some instructors consider it neither continuous nor discontinuous there, just outside the domain.
When making a piecewise function continuous, students sometimes forget to check that c falls in a valid range. If the problem restricts digits to 1 through 9, an algebraic solution of c = 0 would be invalid.
Students assume that a function with a vertical asymptote "equals infinity" at that point. It does not equal anything; it is undefined, and the limit does not exist in the finite sense.
⚠️ Identifying the type of discontinuity (removable, jump, infinite) is a classic short-answer or multiple-choice question. Know the graphical signature and the formal condition for each.
⚠️ "Find the value of c that makes f continuous" is one of the most commonly tested piecewise-function problems in MTH 132. The method is always the same: set the two pieces equal at the boundary.
⚠️ When asked "where is f continuous on [a, b]?", your answer should be a union of intervals. Do not just list the bad points; state the intervals where the function is continuous.
⚠️ Continuity is a prerequisite for the Intermediate Value Theorem. If a problem invokes IVT, the first thing to verify is that f is continuous on the interval in question.
True or False: If lim(x->a) f(x) exists, then f is continuous at a.
False. The function value f(a) must also be defined and equal to the limit.
Fill in the blank: A discontinuity where the left-hand and right-hand limits exist but differ is called a ______ discontinuity.
Jump.
True or False: f(x) = 1/(x - 3) has a removable discontinuity at x = 3.
False. It has an infinite discontinuity (vertical asymptote).
Fill in the blank: To make a piecewise function continuous at x = c, set the two piece formulas equal at x = ______.
c (the boundary point).
Q: Where is f(x) = 1/(1 + sin x) continuous on [0, 2pi]?
A: The denominator is zero when sin x = -1, i.e. x = 3pi/2. So f is continuous on [0, 3pi/2) ∪ (3pi/2, 2pi].
Q: Classify the discontinuity if lim(x->a) f(x) = 5 but f(a) = 3.
A: Removable discontinuity. The limit exists, but f(a) does not equal it.
Q: For f(x) = { x^2 - 10 for x ≤ c ; 8x - 26 for x > c }, find c so that f is continuous everywhere.
A: Set c^2 - 10 = 8c - 26. This gives c^2 - 8c + 16 = 0, so (c - 4)^2 = 0 and c = 4.
Q: A function has a vertical asymptote at x = 2. What type of discontinuity is this?
A: Infinite discontinuity.
Q: Give the three conditions for continuity at a point x = a.
A: (1) f(a) is defined. (2) lim(x->a) f(x) exists. (3) lim(x->a) f(x) = f(a).
Continuity feeds directly into the Intermediate Value Theorem (covered in the companion notes on IVT). IVT requires f to be continuous on a closed interval before it can guarantee a root or a crossing, so every IVT problem starts with a continuity check.
Continuity also connects forward to differentiability: if a function is differentiable at a point, it must be continuous there (but not the other way round). When you study the definition of the derivative, you will see that a discontinuity at a point automatically rules out a derivative there.
continuity, continuous function, discontinuous, removable discontinuity, hole in graph, jump discontinuity, step function, infinite discontinuity, vertical asymptote, piecewise function, piecewise continuity, three conditions for continuity, limit equals function value, domain restriction, denominator equals zero, MTH 132, Calculus I, MSU, Section 1.8