Document

CONTINUITY (PART I)THE CONCEPT OF A CONTINUOUS FUNCTION

In the previous tutorial, we examined various techniques for finding the exact values of limits. One of such

techniques was direct substitution. We saw that most, but NOT ALL limits can be evaluated by simply putting x = a in

the function. In other words, to evaluate the limit

all we have to do is put x = a; that is, evaluate f at a. Functions that can be evaluated this way are said to be

continuous. Thus, by definition, a function f is said to be continuous at a if

Verbally, a function is said to be continuous at a if f(x) approaches f(a) as x approaches a. To help you understand the

concept of a continuous function, we'll tackle it from a geometrical perspective. Take a good look at this graph:

You'll notice from the graph that the curve has no hole or break in it. This is an indication that the function is defined

for all values of x in its domain, Furthermore, this curve can be drawn without removing your pencil from the paper.

A curve with the attributes described above is said to be continuous. The graphs below (next page), however, display

a sharp contrast with the first.

You'll notice a hole on the y-axis (in the second graph) and a break in the curve (on the third graph). These are clear

indications of a discontinuity. As a rule, if you see a hole or a break in any curve, then the function is definitely

undefined at that point, which in turn implies that the function is discontinuous at that point. For a function to be

called “continuous at 'a'”, three conditions MUST be fully satisfied:

CONDITION 1:

which means f must be defined on an interval that contains a.

f(x) limx→a

f(x) = f(a) lim x→a

(i)

f(x) limx→a

MUST EXIST

CONDITION 2:

a must be in the domain of f; that is, f(a) must be defined.

CONDITION 3:

This is another break in the curve.

This hole is an indication of a discontinuity.

The curve breaks at this point;therefore, it is said to be discontinuous

f(x) = f(a) lim x→a

DETECTING DISCONTINUTIESBelow is a graph of a typical function f:

The graph above is that of the piecewise defined function

If we read the graph from the left, we see that there is a discontinuity at a = 0. There are two reasons for this

discontinuity; first, the function is undefined at that point, as indicated by the hole/break. The second reason is that

This is because

Since the left and right hand limits are not equal, then a limit does not exist at a = 0. This means that f is

discontinuous at a = 0.

The purpose of this example is to demonstrate the fact that a particular discontinuity can have more than one cause.

There are other possible discontinuity scenarios:

➔ f(x) may be defined, but limx→a f(x) does not exist (because the left and right limits are not the same);

➔ f(x) may be defined, and limx→a f(x) exists, but f(x) ≠ limx→a f(x).

In the end, we see that a function is discontinuous at a if at least one of the three conditions described above is not

satisfied.

f(x) =

√–x if x < 0

3 – x if 0 ≤ x ≤ 3

(x – 3)2 if x > 3

There is a discontinuity at this point.

f(x) limX→0

DOES NOT EXIST

f(x) = 0 limX→0–

and f(x) = 3 limX→0+

We've seen how we can use a graph of a function to detect a discontinuity. Now, let's try detecting discontinuities

when such functions are mathematically defined (i.e an functions defined by equations/formulas). For each function,

we'll determine where discontinuities occur, if any.

Example 1

SolutionWe see that f is clearly a rational function, and we know that a rational function is undefined when the denominator

evaluates to zero. Thus, to find the values of x that make f undefined (and therefore discontinuous), we are to look for

a value of x such that (x – 1)2 = 0.

We find that x = 1 satisfies the equation above. This means f is undefined when x = 1. Consequently, f is said to be

discontinuous at 1. The graph of f below clearly illustrates the discontinuity. Observe how the curve breaks infinitely

on either side of x = 1:

Based on the nature of the discontinuity, i.e, the way the curve “shoots upward” infinitely, can you guess a suitable

name for this kind of discontinuity?

In the next set of examples, we look at piecewise defined functions.

Example 2

f(x) = 1(x – 1)2

f(x) =if x ≠ 1

2 if x = 1

1x – 1

SolutionThe function f exists for two different domains:

CASE 1: f(x) = 1/(x – 1) if x ≠ 1

CASE 2: f(x) = 2 if x = 1

Obviously, f(1) is defined since f(1) = 2. Ordinarily, f is undefined at x = 1 (from case 1), but the function has been

redefined in case 2. Thus, graphing the function f gives this:

From the graph, you'll see that the curve breaks on either side of x = 1. This is an indication that f is discontinuous at

1. However, this observation is based on the graph. How can we explain this discontinuity mathematically?

Well, as I mentioned earlier, f(1) is defined (and is equal to 2), but

DOES NOT EXIST, because

As you can see, the left and right hand limits are different. Thus, the limit does not exist, and this means that f is

discontinuous at 1. In other words, even though f(1) is defined, the limit

does not exist. Therefore, f is said to be discontinuous at 1.

This function exhibits two kinds of discontinuities: a removable discontinuity and an infinite discontinuity. These will

be explained later on.

f is redefined at this point

f(x) lim x→1

f(x) = –∞ lim x→1–

and f(x) = ∞ lim x→1+

f(x) lim x→1

Example 3

SolutionHere, f has two different domains, like example 2:

CASE 1: f(x) = (x2 – 1)/(x + 1) if x ≠ –1

CASE 2: f(x) = 6 if x = –1

So, f(–1) = 6 is defined (case 1). But as for case 2, f(–1) is undefined. However it does not matter, as f has been

redefined. In this case, we're experiencing a relatively different kind of discontinuity. We see that f(–1) exists, and

exists as well. This is shown in the graph below:

From the graph

but f(–1) = 6. In other words, we are saying that

This inequality indicates that f is discontinuous at –1.

Examples 2 and 3 show relatively similar discontinuities, but understand the basic difference:

➔ For Example 2, the reason for the discontinuity is that CONDITION 1 was not satisfied; whereas, for

Example 3 the reason for the discontinuity is that CONDITION 3 was not satisfied.

f(x) =if x ≠ –1

6 if x = –1

x2 – 1 x +1

f(x) lim x→ –1

f is undefined here

f is redefined here

f(x) = –2 lim x→ –1

f(x) lim x→ –1

≠ f(–1)

Example 4

SolutionGraphing f gives

From the graph, f(2) = –1, which means that f(2) is defined. However, we also see that

which means that

DOES NOT EXIST. In other words, we're saying that f(2) exists, but the limit does not. Thus CONDITION 1 was not

satisfied and therefore, f is discontinuous at 2.

TYPES OF DISCONTINUTIESThere are generally three main types of discontinuities. Understand that a given function may have more than one

type of discontinuity, depending on the nature of the function.

✔ Removable Discontinuity

✔ Infinite Discontinuity

✔ Jump Discontinuity.

f(x) = 1 – x if x ≤ 2

x2 – 2x if x > 2

f(x) = –1 lim x→2–

and f(x) = 0lim x→2+

f(x) lim x→2

Removable DiscontinuityAs the name implies, a removable discontinuity is one that can be eliminated. This is done by simply redefining f at

another value of x = a. This is exactly the type of discontinuity encountered in Examples 2 and 3. Here's another

example:

Observe the hole at point (1,1). There is definitely a discontinuity at this point, because f(1) is not defined. However,

this discontinuity can be eliminated by simply redefining the function f at another number. In this case, the function

has been redefined at the point (1, ½). Understand that you can redefine the function at any point on the graph. So,

we say that the discontinuity has been removed, thus making it a “removable discontinuity”.

Infinite DiscontinuityGenerally, a function whose limit is infinite as x approaches a is said to have an infinite discontinuity at a. In other

words, an infinite discontinuity generally occurs where an infinite limit exists. Here's an example:

f is undefined here

f is redefined here

The graph above clearly shows that

which means

In this kind of situation, we have what is called an infinite discontinuity. Thus f is discontinuous at 1. This kind of

discontinuity occurs in examples 1 and 2 above. Like I mentioned earlier, a function can have more than one type of

discontinuity. In the case of Example 2, there a two discontinuities present: a removable and infinite discontinuity.

Here's another example. This one is crawling with infinite discontinuities:

f(x) = tan(2sin x)

Jump DiscontinuityA function is said to have a jump discontinuity if the function “jumps” from one value to another. An excellent

example is the greatest integer function.

f(x) = –∞ lim x→1–

and f(x) = –∞lim x→1+

f(x) = –∞ lim x→1

Remember from the previous tutorial, that

For this very reason, the greatest integer function is discontinuous at all integers.

In the previous section, we treated an example where a patient was given an injection every four hours. There was a

graph that showed the amount f(t) of drug in the bloodstream after t hours. Does this sound familiar? Well, here's the

graph:

This is another function that exhibits a jump discontinuity.

We have treated three types of discontinuities. Make sure you study each carefully, and know how to identify them,

wherever they occur.

Next, we examine “One-sided continuity”.

ONE–SIDED CONTINUITY If you are reasonably familiar with one sided limits, then this topic shouldn't be terribly difficult. From the definition

of a limit, we saw that

exists, if and only if

exist and are equal. A similar concept applies to one sided continuity.

Let's assume that f(a) exists, then, from one of the conditions of continuity,

which means

Breaking (ii) into two parts gives

[x] = nlim x→a+

[x] = n –1 lim x→a–

and

f(x) lim x→a

f(x) lim x→a+

and f(x) lim x→a–

f(x) lim x→a+

= f(x) lim x→a–

f(x) = f(a) lim x→a

(i)

= f(a) (ii)

So therefore,

➔ If f satisfies (iii), then it is said to be continuous from the right at a;

➔ If f satisfies (iv), then it is said to be continuous from the left at a;

Here's a more formal definition:

LEFT-HAND CONTINUITY:

If

then f is said to be continuous from the left at the number a.

RIGHT-HAND CONTINUITY:

If

then f is said to be continuous from the right at the number a.

To explain the concept of one-sided continuity, we'll use the greatest integer function, f(x) = [x]. From the graph of

this function (click here to view), we see that, for any integer n,

Thus,

This is an indication that f is discontinuous from the left. On the other hand

This means that f is continuous from the right.

Since the greatest integer function is continuous from the right, but discontinuous from the left (for any integer n),

this function is therefore discontinuous at all integers.

Here's another example:

Example 5Find the points at which f is discontinuous. At which of these points is f continuous from the left, from the right, or

neither? Sketch the graph of f.

f(x) lim x→a+

= f(a) (iii)

f(x) lim x→a–

= f(a) (iv)

f(x) lim x→a–

= f(a)

f(x) lim x→a+

= f(a)

f(x) lim x→n–

= f(n)[x] lim x→n–

= n – 1 ≠

f(x) lim x→n–

f(n)≠

f(x) lim x→n+

= f(n)[x] lim x→n+

= n =

f(x) = (x – 1)3 if x < 0

(x + 1)3 if x ≥ 0

SolutionHint: For problems like this one, it is generally helpful to graph the function first. It gives you an idea of what you're

dealing with.

To start with, we graph f:

From f and its graph above, we see that f(0) = 1.

So, we know that f(0) is defined. The next step is to determine the points of continuity on the graph. From the graph,

we see that the most obvious point of discontinuity is at x = 0. This is where we perform the continuity test. We start

by determining the values of the one-sided limits:

For x < 0, we have

For x > 0, we have

From (i) and (ii), we can emphatically say that

and therefore, f is discontinuous at 0. But that's not all.

From the concept of one-sided continuity, we see from (i) that

which means that f is discontinuous from the left at 0. However, (ii) shows that

This implies that f is continuous from the right at 0. Therefore, given the function f, we see that it is discontinuous,

but continuous only from the right, at 0.

f(x) lim x→0–

= f(0) (x – 1)3 lim x→0–

= (0 – 1)3 ≠= –1

f(x) lim x→0+

= f(0) (x + 1)3 lim x→0+

= (0 + 1)3 == 1

(i)

(ii)

f(x) lim x→0

DOES NOT EXIST

f(x) lim x→0–

f(0)≠ (iii)

f(x) lim x→0+

f(0)= (iv)

Example 6Find the points at which f is discontinuous. At which of these points is f continuous from the left, from the right, or

neither? Sketch the graph of f.

SolutionFirst, we graph f:

From the graph, we can establish that f(–1) = –1 and f(1) = 1.

Also, from the graph it is obvious and quite reasonable to assume that there are points of discontinuity: at –1 and 1.

So, let's verify the assumption!!

For x = –1,

This means that f is continuous from the left at –1. On the other hand,

This means that f is discontinuous from the right at –1.

For x = 1,

which implies that f is discontinuous from the left at 1. Conversely,

f(x) =

2x + 1 if x ≤ –1

3x if –1 < x < 1

2x – 1 if x ≥ 1

f(x) lim x→ –1–

= f(–1) (2x + 1) limx→ –1–

= 2(–1) + 1 == –1

f(x) lim x→ –1+

= f(–1) (3x) limx→ –1+

= 3(–1) ≠= –3

f(x) lim x→ 1–

= f(1) (3x) limx→ 1–

= 3(1) ≠= 3

f(x) lim x→ 1+

= f(1) (2x – 1) limx→ 1+

= 2(1) – 1 == 1

This means that f is continuous from the right at 1. In summary,

✔ f is continuous from the left but discontinuous from the right at –1.

✔ f is continuous from the right but discontinuous from the left at 1.

✔ The points of discontinuity of f are therefore –1 and 1, as the graph clearly shows.

Example 7Let

For each of the numbers 2, 3 and 4, discover whether g is continuous from the left, continuous from the right, or

continuous from the number. Sketch the graph of g.

SolutionGraphing g gives

From the function g and its graph above, we see that g(0) = 0, g(2) = 0, g(3) = –1, and g(4) = π.

To find how g is continuous at the numbers 2, 3 and 4, evaluate the left and right hand limits of g at each of the

numbers.

For x = 2,

and

Since g is continuous from the left and from the right at 2, it is therefore continuous at 2.

g(x) =

2x – x2 if 0 ≤ x ≤ 2

2 – x if 2 < x ≤ 3 x – 4 if 2 < x < 3

π if x ≥ 4

g(x) lim x→2–

= g(2) (2x – x2) limx→2–

= 2(2) – (2)2 == 0

g(x) lim x→2+

= g(2) (2 – x) limx→2+

= 2 – (2) == 0

For x = 3,

and

Again, we find that g is is continuous from either side at 3, which means g is continuous at 3.

For x = 4,

and

Thus, at the number 4, g is discontinuous from the left, but continuous from the right.

We've looked at how we can determine whether a function is continuous or not at a particular number a. But if given a function f, how can we really prove that it is actually continuous on a specific interval, or at a number? We'll answer this question in the following section, after which we'll examine “Continuity Theorems”.

PROVING CONTINUITY Let's take a simple function, say, f(x) = 2x2 – 5, and let's pick a relatively small interval [–3, 4]. The question here is,

is f defined for every number in the interval [–3, 4]? Of course it is, and since the three conditions of continuity have

been fully satisfied, we say that f is continuous on the interval. So, by definition

If a function f is continuous on EVERY number in an interval [a, b], where a

and b are the endpoints of the interval, then it is continuous on that interval.

Note that this definition is for continuity on an interval. For continuity at a number,

If f is defined on only one side of an endpoint of an interval, we understand continuous

at the endpoint to mean continuous from the left or continuous from the right.

Simply put, to prove that a function is continuous at a number or on an interval, we need to prove that

In the following examples, we will use the definition of continuity and the properties of limits to show that the given

functions are continuous on the given intervals/at the given numbers.

Example 8f(x) = x2 + √7 – x a = 4

SolutionIf f were to be continuous at 4, it would mean that

g(x) lim x→3–

= g(3) (2 – x) limx→3–

= 2 – (3) == –1

g(x) lim x→3+

= g(3) (x – 4) limx→3+

= (3) – 4 == –1

g(x) lim x→4–

= g(4) (x – 4) limx→4–

= (4) – 4 ≠= 0

g(x) lim x→4+

= g(4)π limx→4+

= π =

f(x) = f(a) lim x→a

f(x) = f(4) lim x→4

So, here's what we do:

So,

Since f(x) = x2 + √7 – x

then, putting x = 4 gives

f(4) = (4)2 + √7 – (4)

f(4) = 16 + √3

From (i) and (ii), we see that

And therefore, we say that f is continuous at 4.

Hint: Graphing the function is a good way of verifying the result.

Do you see how it works? Let's see more examples.

Example 9f(x) = (x + 2x3)4 a = –1

SolutionIf f is continuous at –1, then it would mean that

So therefore,

x2 + √7 – x f(x) lim x→4

= lim x→4

= x2 lim x→4

+ lim x→4 √7 – x Addition law

= x2 lim x→4

+ – 7 lim x→4

x lim x→4 √

Addition law

Root law, Subtraction law

= (4)2 + √7 – (4)

= 16 + √3

f(x) lim x→4 = 16 + √3 (i)

(ii)

f(x) = f(4) lim x→4

f(x) = f(–1) lim x→–1

f(x) lim x→–1

= (x + 2x3)4 lim x→–1

= (x + 2x3) lim x→–1

4Power law

= x lim x→–1

x3 lim x→–1 2+

4Addition/Constant Multiple law

= (–1 + 2(–1)3)4

= 81

Incidentally, we also find that

f(–1) = ((–1) + 2(–1)3)4 = 81

Thus, we see that

which is clear proof that f is continuous at –1.

Example 10

SolutionIf f is continuous at 4, then

So, this time, let's evaluate both sides of the equation simultaneously:

Therefore, since

then g is continuous at 4.

Example 11

f(x) = x√16 – x2 [–4, 4]

SolutionLet's pick a number a in the interval [–4, 4] . In other words, we pick a number a such that –4 < a < 4. We want to

prove that f is continuous on the interval [–4, 4]. It is therefore expected that f will be continuous at every number a

in this interval, which is why we pick a number a in the interval to prove the continuity. So, given the function f and

f(x) = f(–1) lim x→–1

g(x) =x + 1

2x2 – 1 a = 4

g(x) = f(4) lim x→4

lim x→4

=x + 1

2x2 – 1

(4) + 1

2(4)2 – 1

x lim x→4

1lim x→4

+

x2lim x→4 2 – 1lim

x→4

=5

31

(x + 1)lim x→4

(2x2 – 1)lim x→4

=

= =5

31

=(4) + 1

2(4)2 – 1 =

5

31

= 5

31=

5

31

g(x) = f(4) lim x→4

the number a, we are to prove that

from the definition of continuity. So, here goes:

The result above shows that f is continuous at a if –4 < a < 4. In other words, f is continuous on the interval [–4, 4].

This is illustrated in the graph of f:

Example 12Use continuity to evaluate the limit

f(x) = f(a) lim x→a

x√16 – x2 f(x) lim x→a

= lim x→a

= x lim x→a

lim x→a

√16 – x2 Product law

= x lim x→a

– 16 lim x→a

x2 lim x→a √

(16 – x2) lim x→a √

= x lim x→a

Root law

Subtraction law

a√16 – a2 =

= f(a)

lim x→4

5 + √x

√5 + x

SolutionHere, we simply apply the basic principle of continuity:

So, let's assume

This means that if f were truly continuous at 4, then all we need to do to evaluate the limit is simply substitute

directly, that is, evaluate f at 4, which gives

Therefore,

Recall that we used this technique (direct substitution) extensively in solving limits in the previous tutorial.

Example 13For what value of the constant c is the function f continuous on (–∞, ∞)?

SolutionTo tackle this problem, we go back to the basics; that is, we employ the number one rule of continuity, which in this

case, is

From f, we see that f(3) = 3c + 1.

Also, we find that

and

We have a problem here: the left and right hand limits are not equal, which would mean that f is discontinuous at 3.

BUT WAIT!!

Recall that, for a function to be continuous at a number a, both left and right hand limits have to be EQUAL. So,

assuming the two limits are equal, then we say

Which gives

f(x) = f(a) lim x→a

5 + √x

√5 + x f(x) =

lim x→4

5 + √x

√5 + x =

5 + √4

√5 + 4 =

5 + 2

√9 =

7

3

lim x→4

5 + √x

√5 + x =

7

3

f(x) = cx + 1 if x ≤ 3

cx2 – 1 if x > 3

f(x)lim x→a–

= f(x)lim x→a+

= f(a)

f(x)lim x→3– = (cx + 1)lim

x→3– = 3c + 1 = f(3)

f(x)lim x→3+ = (cx2 – 1 )lim

x→3+ = 9c – 1 ≠ f(3)

f(x)lim x→a–

= f(x)lim x→a+

3c + 1 9c – 1 =

To find what makes these two limits equal, we solve for c:

3c + 1 = 9c – 1

3c – 9c = – 1 – 1

c = 1/3

In other words, by putting c = 1/3, the left and right hand limits would be equal and thus f will be

continuous at 3, and ultimately, this means that f is continuous at on (–∞, ∞), since f is made up of

polynomials. The graph of f is shown below:

Example 14Find the constant c that makes g continuous on (–∞, ∞)

SolutionAgain, we apply the basic principle of continuity:

We see that g(4) = 4c + 20.

Also, we find that

and

Like example 13 above, we are faced with the very same problem: left and right hand limits are not equal. Thus, we

solve for the common factor c which binds both limits together:

g(x) = x2 – c2 if x < 4

cx + 20 if x ≥ 4

g(x)lim x→a– = g(x)lim

x→a+ = g(a)

g(x)lim x→4–

= (x2 – c2)lim x→4–

= (4)2 – c2 = g(4)16 – c2 ≠

g(x)lim x→4+

= (cx + 20)lim x→4+

= (4)c + 20 = g(4)4c + 20 =

16 – c2 = 4c + 20

Rearranging the equation gives

– c2 – 4c + 16 – 20 = 0

= – c2 – 4c – 4 = 0

Using the general quadratic formula [1] to solve for c, we find that c = –2 .

Therefore, the ONLY value of c for which g is continuous on (–∞, ∞) is –2. Again, g is continuous on this interval

because it is made up of polynomial functions. g is graphed below:

[1] NOTE:

The general quadratic formula is

where a, b and c are coefficients in the general quadratic equation

ax2 + bx + c

Example 15If

show that f is continuous on (–∞, ∞).

SolutionThere are two ways of proving that f is continuous for all x:

(1) Mathematical proof:

By definition, the functions that make up f, that is, the functions y = x – 1 and y = 5 – x are linear polynomials (since

they are of degree 1). We also know that the domain of ANY polynomial is (–∞, ∞), which means a polynomial is

defined for EVERY number a.

– b ± √b2 – 4ac

2ax =

f(x) = x – 1 if x < 3

5 – x if x ≥ 3

Now that these basic facts have been established, we can easily prove the continuity of f. To do this, we simply need

to show that

Specifically, we need to show that f is continuous at 3. In other words, we only need to prove this:

First, we understand that f(3) = 2.

As for the limits, we have

and

Clearly, we see that

Thus, we can categorically say that f is continuous at 3. This result also signifies that f is continuous on (–∞, ∞).

Based on how f was defined, we know for sure that it is defined FOR ALL x. However, it is likely that there would be

a discontinuity at 3. Since we have proven that there isn't any discontinuity, this means f is indeed defined FOR ALL

x.

(2) Graphical proof: this simply complements the mathematical proof :

Example 15If f and g are continuous functions with f(3) = 5, and

find g(3).

f(x) = f(a) lim x→a

f(x)lim x→3–

= f(x)lim x→3+

= f(3)

f(x)lim x→3–

= (x – 1)lim x→3–

= (3) – 1 = f(3)2 =

f(x)lim x→3+

= (5 – x)lim x→3+

= 5 – (3) = f(3)2 =

f(x) = f(3) lim x→3

[2f(x) – g(x)] = 4lim x→3

SolutionUsing the difference and constant multiple laws of limits, we have

which means

Since f(3) = 5 and f is continuous, then it means

Therefore, we have

And so,

Hence,

Since g is also continuous, then from the definition of continuity,

which means that g(3) = 6.

As you can see, tackling this question requires conceptual understanding of continuity, that is, an

understanding of the fundamental principles of continuity.

In Part II of this tutorial, we take a look at continuity theorems, which will help ease the task of solving problems

involving continuity, and, we examine the Intermediate Value Theorem.

calculus4engineeringstudents.com

[2f(x) – g(x)] lim x→3

= f(x) lim x→3

2 – g(x) lim x→3

f(x) lim x→3

2 – g(x) lim x→3

= 4

f(x) = f(3) = 5 lim x→3

2(5) – g(x) lim x→3

= 4

10 – g(x) lim x→3

= 4=

g(x) lim x→3

= 10 – 4

= 6

g(x) lim x→3

= 6

g(x) = g(3) = 6 lim x→3