: At a Glance

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Topics At a Glance
Continuity at a Point
Continuity at a Point via Pictures
Continuity at a Point via Formulas
Functions and Combinations of Functions
Continuity on an Interval
Continuity on an Interval via Pictures
Continuity on an Interval via Formulas
Continuity on Closed and Half-Closed Intervals
Determining Continuity
The Informal Version
The Formal Version
Properties of Continuous Functions
Boundedness
Extreme Value Theorem
Intermediate Value Theorem
In the Real World
Page: I Like Abstract Stuff; Why Should I Care?
Page: How to Solve a Math Problem
Appendix: Intervals and Interval Notation

Introduction to :

It's good to have a feel for what continuity at a point looks like in pictures. However, sometimes we are asked about the continuity of a function for which we're given a formula, instead of a picture. When this happens, remember that the following three statements must all hold for f to be continuous at c.

I. The function f is defined at x = c.
The limit exists.
The value f(c) agrees with the limit .

Practice:

Examples
Exercises

Example 1

Determine whether the function

is continuous at x = 1.

Make sure the function is defined at x = 1. It is, and f(1)=1.

Now we will see if exists. As x approaches 1 from the left, we find

Similarly, as x approaches 1 from the right we find

Since the one-sided limits agree,

Since , the function f is continuous at x = 1.

Example 2

Determine whether the function

is continuous at x = 2.

When we try to evaluate the function at x = 2, we find
which is undefined! Since f(2) doesn't exist, the function is not continuous at x = 2.

Example 3

Determine whether the function

is continuous at x = 0.

When we evaluate the function at 0 we find

which is fine.

We will see if the limit of f exists as x approaches 0. As x approaches 0 from the left, x will be less than 0. We use the first part of the function definition and find

As x approaches 0 from the right, x will be greater than 0. We need to use the second line of the function definition. We find

Since the one-sided limits do not agree, does not exist. The function f is not continuous at x = 0.

Example 4

Determine whether the function

is continuous at x = 0.

This function is not continuous at x = 0. The function is defined for x<0 and x>0, but not for x = 0, Therefore f(0) is undefined.

Further, is undefined since

Example 5

At what values is the function f discontinuous?

The function f will be discontinuous at any value of c where f(c) is undefined, where does not exist, or where . Think of such values as problem spots. We need to look at the function definition and find these "problem spots".

Where is f undefined?
It looks like x = 2 could be trouble, since then

However, when x = 2 we're not using this part of the function definition. We only use when x<0. Since 2 > 1, we have f(2) = 4.

The function f is undefined at . When we're using the piece of the function definition that says

we find

Therefore f is discontinuous at .

Where can not exist?
This can happen in between the "pieces", that is, where the function definition changes - or where individual pieces have undefined limits. The only place an individual piece of this function has an undefined limit is at , which we've already taken care of.

For this function, the function definition changes at 0 and 1. We need to investigate the limit of f as x approaches each of these values. First off, does exist? For the left-sided limit, we find

For the right-sided limit, we find

Since the one-sided limits disagree, does not exist, and f is discontinuous at x = 0.

Does exist?
For the left-sided limit, we find

For the right-sided limit, we find

Since the one-sided limits agree,

Finally, we need to see if f(c) and disagree anywhere. Again, we only need to inspect those values of c where the function definition changes over. Since the function f is continuous at 1.

We have no more possible trouble spots, f is discontinuous only at and x = 0.

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Exercise 1

Determine whether the function

is continuous at each given value. If not, explain.

x = -5
x = -4
x = 0
x = 2
x = 3

Answer

When we try to evaluate f(-5), we find which is undefined. Since f(-5) does not exist, f is discontinuous at x = -5.

When we try to evaluate f(-4), we find f(-4) = 0, which is fine. We will see if exists and agrees with f(-4). As x approaches -4 from the left, we use the part of the function definition that says The left-sided limit is When x is approaches -4 from the right, x is just a bit bigger than -4, we use the part of the function definition that says f(x) = x - 8 for -4 < x ≤ 2. Then the right-sided limit is Since the left- and right-hand limits agree, Alas, since f(-4) = 0, the limit disagrees with the function value, therefore f is discontinuous at x = -4.

f(0) = 0 - 8 = -8. As x approaches 0 from the left or the right, we use the part of the function definition that says f(x) = x-8 for -4the function is continuous at x = 0.

f(2) = 2 - 8 = -6, which is fine. Now we will check the limit. As x approaches 2 from the left we use the part of thefunction definition that says f(x) = x-8 for -4 < x ≤ 2 we find As x approaches 2 from the right we use the part of the function definition that says and from earlier work with limits we know that is undefined. Therefore does not exist, therefore f is discontinuous at x = 2.

f(3) we findwhich is undefined. Therefore f is discontinuous at x = 3.

Exercise 2

For the function, determine all values at which the function is discontinuous.

Answer

f is defined everywhere, there can't be any values of x where f is discontinuous because it is undefined. That means we need to look for places where the limit of f is undefined or disagrees with the value of f. The only place this could possibly happen is x = 0. We will investigate. The left-hand limit is

The right-hand limit is

The one-sided limits agree,

Alas, this disagrees with the function value f(0) = 0. The function is discontinuous at x = 0.

Exercise 3

For the function, determine all values at which the function is discontinuous.

Answer

First, we will look for values of x at which g is undefined. x = 1 is fine, since g(1) = 2 by the second piece of the definition. x = 2 is not fine: the function definition says what should happen for x < 2 and 2 < x, but doesn't make any provision for what happens at x = 2. Therefore g is undefined, and therefore discontinuous, at x = 2.

Now we will look for places where the limit is undefined or disagreeable. Since is undefined, is also undefined and g is discontinuous when x = 1.

To summarize, the function g is discontinuous at x = 1 and x = 2.

Exercise 4

For the function, determine all values at which the function is discontinuous.

Answer

The function definition says

The denominator factors, therefore we could rewrite this as

This function is undefined, and therefore discontinuous, at x = 4 and x = 5.The only place limits could go wrong, besides x = 4 and x = 5, is when x = 1. At x = 1 we have

As x approaches 1 from the left we have

As x approaches 1 from the right we have

Since the one-sided limits agree,

This agrees with h(1) = 1,

so the function is continuous at x = 1.

To summarize, h is only discontinuous at x = 4 and x = 5.

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