http://calculus4engineeringstudents.com/Continuity

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

lim f(x)

x→a 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

lim

f(x) = f(a)

(i)

x→a 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:

lim f(x)

x→a

MUST EXIST

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

This hole is an indication of a discontinuity.

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

This is another break in the curve.

CONDITION 2:

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

CONDITION 3:

lim

x→a

f(x) = f(a)

DETECTING DISCONTINUTIES Below is a graph of a typical function f:

There is a discontinuity at this point.

The graph above is that of the piecewise defined function

√–x f(x) =

if x < 0

3–x

if 0 ≤ x ≤ 3

(x – 3)2

if x > 3

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

lim f(x) DOES NOT EXIST

X→0 This is because

lim f(x) = 0

X→0 –

and

lim f(x) = 3

X→0 +

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.

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 f(x) =

1 (x – 1)2

Solution We 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

if x ≠ 1 if x = 1

Solution The 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:

f is redefined at this point

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

lim f(x) x→1

DOES NOT EXIST, because

lim f(x) = –∞

x→1 –

and

lim f(x) = ∞

x→1 +

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

lim f(x)

x→1 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.

Example 3 x2 – 1 x +1 6

f(x) =

if x ≠ –1 if x = –1

Solution Here, 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

lim f(x) x→ –1

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

f is redefined here

f is undefined here

From the graph

lim

f(x) = –2

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

lim f(x)

≠ f(–1)

x→ –1 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.

Example 4 f(x) =

1–x

if x ≤ 2

x2 – 2x

if x > 2

Solution Graphing f gives

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

lim

x→2 –

f(x) = –1

and

lim f(x) = 0

x→2 +

which means that

lim f(x) x→2

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 DISCONTINUTIES There 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.

Removable Discontinuity As 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:

f is undefined here f is redefined here

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 â&#x20AC;&#x153;removable discontinuityâ&#x20AC;?.

Infinite Discontinuity Generally, 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:

The graph above clearly shows that

lim f(x) = –∞

x→1 –

and

lim f(x) = –∞

x→1 +

which means

lim f(x) = –∞

x→1 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 Discontinuity A 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.

Remember from the previous tutorial, that

lim [x] = n lim [x] = n –1 and x→a+ x→a– 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

lim f(x)

x→a exists, if and only if

lim f(x)

and

x→a+

lim f(x)

x→a–

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,

lim f(x) = f(a)

x→a

(i)

which means

lim f(x)

x→a+

Breaking (ii) into two parts gives

=

lim f(x)

x→a–

=

f(a)

(ii)

lim f(x)

=

x→a+

lim f(x)

=

x→a–

f(a)

(iii)

f(a)

(iv)

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

lim f(x)

=

x→a–

f(a)

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

RIGHT-HAND CONTINUITY: If

lim f(x)

=

x→a+

f(a)

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,

lim f(x)

x→n–

=

lim [x]

x→n–

=

n–1

≠

f(n)

Thus,

lim f(x)

x→n–

≠

f(n)

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

lim f(x)

x→n+

=

lim [x]

x→n+

=

n

=

f(n)

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 5 Find 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) =

(x – 1)3

if x < 0

(x + 1)3

if x ≥ 0

Solution Hint: 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

lim f(x)

=

lim f(x)

=

x→0 –

lim (x – 1)3

=

(0 – 1)3

=

–1

≠

f(0)

(i)

lim (x + 1)3

=

(0 + 1)3

=

1

=

f(0)

(ii)

x→0 –

For x > 0, we have x→0 +

x→0 +

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

lim f(x)

DOES NOT EXIST

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

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

lim f(x)

x→0 –

≠

f(0)

(iii)

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

lim f(x) = f(0) (iv) x→0 + 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.

Example 6 Find 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.

2x + 1

if x ≤ –1

3x

if –1 < x < 1

f(x) =

2x – 1

if x ≥ 1

Solution First, 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,

lim f(x)

x→ –1 –

=

lim (2x + 1)

x→ –1 –

=

2(–1) + 1 =

–1

=

f(–1)

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

lim f(x)

x→ –1 +

=

lim (3x)

x→ –1 +

=

3(–1)

=

–3

≠

f(–1)

=

3

≠

f(1)

1

=

f(1)

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

For x = 1,

lim

x→ 1 –

f(x)

=

lim (3x)

x→ 1 –

=

3(1)

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

lim

x→ 1 +

f(x)

=

lim (2x – 1)

x→ 1 +

= 2(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 7 Let

g(x) =

2x – x2

if 0 ≤ x ≤ 2

2–x x–4

if 2 < x ≤ 3 if 2 < x < 3 if x ≥ 4

π

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.

Solution Graphing 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,

lim g(x)

=

lim g(x)

=

x→2 –

lim (2x – x2)

=

2(2) – (2)2

=

0

=

g(2)

lim (2 – x)

=

2 – (2)

=

0

=

g(2)

x→2 –

and x→2 +

x→2 +

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

For x = 3,

lim g(x)

=

lim g(x)

=

x→3 –

lim (2 – x)

=

2 – (3)

=

–1

=

g(3)

lim (x – 4)

=

(3) – 4

=

–1

=

g(3)

x→3 –

and x→3 +

x→3 +

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

For x = 4,

lim g(x)

=

lim g(x)

=

x→4 –

lim (x – 4)

=

x→4 –

(4) – 4

=

0

≠

g(4)

and x→4 +

lim

x→4 +

π

=

π

=

g(4)

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) = 2x 2 – 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

lim f(x) = f(a) x→a

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 8 f(x) = x2 +

√7 – x

a = 4

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

lim f(x) = f(4) x→4

So, here's what we do:

lim

f(x)

x→4

x2 + √7 – x

=

lim

=

lim x2

+

lim

=

lim x2

+

√

=

x→4

x→4 x→4

(4)2 +

=

√7 – x

x→4

lim 7 x→4

Addition law

–

lim x x→4

Root law, Subtraction law

√7 – (4)

16 +

√3

So,

lim

x→4

Since f(x) = x2 +

f(x)

=

16 +

√3

(i)

√7 – x

then, putting x = 4 gives f(4) = (4)2 + f(4) = 16 +

√7 – (4)

√3

(ii)

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

lim f(x) = f(4) x→4

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 9 f(x) = (x + 2x3)4

a = –1

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

lim

x→–1

f(x) = f(–1)

So therefore,

lim f(x)

x→–1

=

=

=

lim (x + 2x3)4

x→–1

lim (x + 2x3) 4

Power law

x→–1

lim

x→–1

=

(–1

=

81

x

+

+ 2(–1)3)4

2 lim x3 x→–1

4

Addition/Constant Multiple law

Incidentally, we also find that

f(–1) = ((–1) + 2(–1)3)4 = 81 Thus, we see that

lim

x→–1

f(x) = f(–1)

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

Example 10 x+1

g(x) =

a = 4

2x2 – 1

Solution If f is continuous at 4, then

lim g(x) = f(4) x→4

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

lim

x→4

x+1 2x2 – 1

=

(4) + 1 2(4)2 – 1

=

5 31

lim (x + 1) =

x→4

lim (2x2 – 1) x→4

=

lim x

lim 1

+

x→4

2 lim x2 x→4 =

(4) + 1 2(4)2 – 1

=

5 31

x→4

lim 1

–

5 31

x→4

5 31

= =

=

5 31

Therefore, since

lim g(x) = f(4) x→4

then g is continuous at 4.

Example 11 f(x) = x√16 – x2

[–4, 4]

Solution Let'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

the number a, we are to prove that

lim f(x) = f(a) x→a

from the definition of continuity. So, here goes:

lim

x→a

f(x)

=

=

=

= = =

lim

x→a

x√16 – x2

lim x

x→a

lim x

x→a

lim x

x→a

lim √16 – x2

Product law

x→a

√ √

lim (16 – x2)

Root law

x→a

lim 16

x→a

–

lim x2 x→a

Subtraction law

a√16 – a2 f(a)

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 12 Use continuity to evaluate the limit

lim

x→4

5 + √x

√5 + x

Solution Here, we simply apply the basic principle of continuity:

lim f(x) = f(a) x→a

So, let's assume

f(x) =

5 + √x

√5 + x

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

5 + √x

lim

x→4

5 + √4

=

√5 + x

=

√5 + 4

5+2

7 3

=

√9

Therefore,

5 + √x

lim

7 3

=

√5 + x

x→4

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

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

f(x) =

cx + 1

if x ≤ 3

cx – 1

if x > 3

2

∞)?

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

lim f(x)

x→a–

From f, we see that f(3)

=

lim f(x)

=

x→a+

f(a)

= 3c + 1.

Also, we find that

lim f(x)

=

lim f(x)

=

x→3–

lim (cx + 1)

=

3c + 1

=

f(3)

lim (cx2 – 1 )

=

9c – 1

≠

f(3)

x→3–

and

x→3+

x→3+

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

lim f(x)

=

3c + 1

=

x→a–

lim f(x)

x→a+

Which gives

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

g(x) =

∞)

x2 – c2

if x < 4

cx + 20

if x ≥ 4

Solution Again, we apply the basic principle of continuity:

lim g(x)

x→a–

lim g(x)

=

g(a)

lim (x2 – c2)

=

(4)2 – c2

lim (cx + 20)

=

=

x→a+

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

Also, we find that

lim g(x)

=

lim g(x)

=

x→4–

x→4–

=

16 – c2

≠

g(4)

4c + 20

=

g(4)

and

x→4+

x→4+

(4)c + 20 =

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:

16 – c2 = 4c + 20 Rearranging the equation gives

– c2 – 4c + 16 – 20 2

=

– c – 4c – 4

Using the general quadratic formula

[1]

= 0

= 0

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

x =

± √b2 – 4ac

–b

2a

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

ax2 + bx + c

Example 15 If

f(x) =

x–1

if x < 3

5–x

if x ≥ 3

show that f is continuous on (–∞, ∞).

Solution There 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.

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

lim f(x) = f(a) x→a

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

lim f(x)

x→3–

=

lim f(x)

x→3+

=

f(3)

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

As for the limits, we have

lim f(x)

=

lim f(x)

=

x→3–

lim

(x – 1)

=

(3) – 1

=

2

=

f(3)

lim

(5 – x)

=

5 – (3)

=

2

=

f(3)

x→3–

and

x→3+

x→3+

Clearly, we see that

lim f(x) = f(3) x→3

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

lim

x→3

find g(3).

[2f(x)

]

– g(x)

= 4

Solution Using the difference and constant multiple laws of limits, we have

lim

x→3

[2f(x)

]

=

– g(x)

2 lim f(x) x→3

–

lim g(x)

x→3

which means

2 lim f(x) x→3

lim g(x)

–

x→3

=

4

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

lim f(x)

x→3

=

f(3)

= 5

Therefore, we have

=

2(5)

–

10

–

lim g(x)

=

4

lim g(x)

=

4

x→3

x→3

And so,

lim g(x)

x→3

=

10 – 4

=

6

Hence,

lim g(x)

x→3

=

6

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

lim g(x)

x→3

=

g(3)

= 6

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.

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