The topic of this post is continuity. The phrase “a function is continuous on its domain” was much discussed last week on the AP Calculus Community bulletin board as it is about this time every year. This led to a discussion of one-sided continuity at the endpoint of an interval.

Let’s start by looking at some definitions related to continuity. A detailed discussion of definitions is worth doing in any math class; it helps students learn about the structure of mathematics. Definitions are biconditional statements; that is, either part implies the other.

**Definitions related to Continuity**

, (a,__A function is continuous at a point__*f*(*a*)) if, and only if, (1)*f*(*a*) exists (is a finite number), and (2) exists (is a finite number), and (3) – the limit equals the value.

This is the three-part definition you will find in most textbooks. The first two conditions are sort of for emphasis since the third is not possible if the limit does not exist (or is infinite) and/or the value does not exist (DNE). Graphically this means that as you move along the graph of *f* getting closer to the place where *x* = *a* the function values are getting closer to *f*(*a*). This is not always the case; specifically, if this does not happen the function is not continuous at the point.

For example, let

- This function is undefined at
*x*= 2 and*x*= 3; at both values*f*(*x*) DNE. - At
*x*= 3 the limit does exist and is equal to 1. In this case the graph appears to be missing a single point, here (3, 1). This is called a*removable discontinuity*because be redefining the function so that*f*(3) = 1 the function could be made continuous. This happens most often when the factor in the denominator that becomes zero “cancels” with a like factor in the numerator. This can also happen if the function value is defined as something different than the limit. There are other examples such as at*x*= 0 - At
*x*= 2 the limit DNE, and the situation appears as a*vertical asymptote*on the graph: on the left the values approach negative infinity and on the right they approach positive infinity. This is because a factor in the denominator gets very small as you approach 2.

Proving a function is continuous has been asked on AP Calculus exams. (See 2003 AB 6a and 2011 AB 6a) Students were expected to state the value of the two one-sided limits and the value of the function to show the three parts of the definition were true.

if, and only if, it is continuous at every point in the interval.__A function is continuous on an open interval__

The definition of continuous at a point is extended to all the points of an interval. But do you have to check each and every point? No, this is determined by finding where the function is not continuous; the function is then continuous on the intervals between these points.

- In many theorems it is necessary to have a function be continuous on a closed interval; the endpoints must be included, but the limit in the main definition is a two-sided limit. The main definition is adapted to read
at__A function is continuous from the right__*x = a*(or from theat__left__*x*=*b*) if, and only if, .

The limits are approached from inside the interval: from the right at the left end and from the left at the right end.

**Continuous on its Domain**

Functions exist only on their domain, so there is no way a function can be continuous except on its domain. I wrote about this is a previous post Right answer, Wrong Question.

When analyzing a new function one of the things you look for is where the function is continuous. But really you do this by looking for places where it is not continuous; everywhere else the function will be continuous.

Vertical asymptotes, removable discontinuities, and oscillating discontinuities are not possible for a function continuous on its domain; these appear on the graph, but mark places where the function DNE. The only kind of discontinuity a function may have *on its domain* are jump or step discontinuities. These often occur with piecewise defined functions where the endpoint of one piece is defined and different from the corresponding endpoint of the next piece. For example,

- Since the function is defined there, but the limits of the two parts (0 and -1) are not the same, so the function is defined, but not continuous on its domain.

Another example is the greatest integer function (aka the floor function) whose graph is shown in below. Note that the left-endpoint of each segment (at the integer values) is included, while the right-endpoint at the next integer is not included.

__Spotting discontinuities__

Here are some hints on looking for places where a function is not continuous:

- Look for the zeros of the denominators
- For piecewise functions check the places where the pieces change definition.
- For trigonometric functions, the sine and cosine are continuous for all
*x.*Since the other trigonometric functions are defined in terms of the sine and cosine look places where the sine or cosine is zero and appears in the denominator of the definition. - Function involving square roots and other even degree roots are not defined, and therefore not continuous where their arguments are negative. For example, is continuous on the closed interval [-2, 2] (using the definitions for closed intervals – one-sided limits), but not elsewhere.
- Use technology to graph the function; discontinuities usually show up. The exception is a removable discontinuity or an oscillating discontinuity which may not show up because the
*x-*value of the discontinuity may not be a pixel coordinate.

**Pathological Examples**

- This function, attributed to Dirichlet, is defined for all Real numbers, but not continuous anywhere.

- Between every pair of irrational numbers is a rational number, and between every pair of rational numbers is an irrational number.

- With a slight variation, we have this function:

- This function is defined for all Real numbers and is continuous at only one point (1, 1). Go figure.

.

why every continuous function is not differentiable?

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Take a very simple counterexample like y = | x |, the absolute value function. This function, and others like it, are not differentiable at (0, 0) because the limit expression for the derivative is not the same approaching zero from the right and left. To the left of the origin the slope (that is, the derivative) is –1; to the right the slope is +1. Therefore, at x = 0 the derivative has a jump discontinuity and does not exist at the origin. For more on this see these post Local Linearity, Local Linearity I,Local linearity II. and Tangent Lines

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I believe the first function was first considered by Dirichlet.

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Jim – Thanks for spotting this. I made the correction.

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