/Continuity(II)

CONTINUITY (PART II) CONTINUITY THEOREMS In the final section if the previous tutorial, we saw how we could prove the continuity of a function. This was accomplished by simply applying the fundamental definition of continuity. In other words, we proved the continuity of a given function by showing that

lim f(x) = f(a)

x→a or

lim f(x) = f(a)

x→a– or

lim f(x) = f(a)

x→a+

as the case may be. Clearly, this method of proving continuity is rather tiresome.

So now, instead of using the definition to prove continuity, we use Theorems. Here, we look at four basic Theorems. In the end, you'll find that this method of proving continuity is much better. We start with Theorem 1:

Theorem 1: If f and g are continuous at a and c is a constant, then the following functions are also continuous at a: (I).

f + g

(II).

f – g

(III).

cf

(IV).

fg

(V).

f/g

{if g ≠ 0}

For your first exercise in this section, try proving Theorem 1 [parts (I) through (V)].

Hint: Use the concept of the limit laws. For example, to prove part (I), use the idea that the limit of a sum is the sum of limits (addition law of limits).

Theorem 2: (I).

A polynomial is continuous everywhere.

(II).

A rational function is continuous wherever it is defined; that is, it is continuous on its domain.

(III).

Polynomials, rational functions, root functions and trigonometric functions are continuous at EVERY number in their domains.

Using Theorems 1 and 2, we can prove the continuity of a function by simplifying it; that is, by expressing the function as a sum, difference, product or quotient. After breaking up the function, then we apply Theorem 2 to prove the continuity of the individual functions. Here's a simplified explanation: Suppose h is a sum of two simpler functions, that is, h(x) = (f + g)(x). The task here is to prove that h is continuous.

Here's what you do: express h as a sum of two functions f and g, and prove that both functions are continuous using the

appropriate Theorems. Since we have shown that f and g are continuous, then consequentially, the sum f + g, will also be continuous, and thus, h is continuous. That's the concept you need to understand.

Here's an example:

Example 1 Explain, using Theorems 1 and 2, why F is continuous at every number in its domain. State its domain.

F(x) =

x x2 + 5x + 6

Solution It is quite obvious that F is a rational function, it takes the form F(x) = (f/g)(x), where f(x) = x g(x) = x2 + 5x + 6

By Theorem 2, both f and g are continuous everywhere, because they are polynomials. Thus, by Part V of Theorem 1, F is continuous for every number in its domain. But the next question is, what is the domain of F? Recall that F is a rational function, which means that under no circumstance must g equal zero., otherwise F is undefined. In other words, F is continuous everywhere except where g = 0. We see that this occurs when x = –2 and –3.

Ultimately, this implies that F is continuous at every number except –2 and –3, and so, its domain is

{x|x ≠ –2, x ≠ –3} Now do you see how the Theorems work? We'll look at more examples later.

Next we have Theorems 3 and 4, which are particularly useful for proving the continuity of composite functions, that is, functions that are formed by composition, rather than by simple combination. To help you understand Theorems 3 and 4 fully, let's briefly discuss composite functions.

COMPOSITE FUNCTIONS Real numbers can be combined by addition, subtraction, multiplication and division. The same applies to functions. In other words, given two functions f and g, we can have f + g, f – g, fg, and f/g. This is the simple method of combining functions. For reference purposes, we'll call this method simple combination.

Alternatively, a new function can be formed by composition. Essentially, this method of combination involves substituting one function into another. By definition,

Given two functions f and g, then the composition of f and g is given by (f ೦ g)(x) = f (g(x)) The function f ೦ g is what we call a COMPOSITE FUNCTION. The process of substituting one function into another is called COMPOSTION. The composite function f ೦ g is so-called because it is composed of two functions: f and g, and is pronounced “f circle g”.

Here's an illustration.

Suppose f(x) = x3 and g(x) = 2 – x, find f ೦ g.

From the definition of a composite function, we know that (f ೦ g)(x) = f (g(x))

Therefore, we substitute g into f to give: f(g(x)) = f(2 – x)

= (2 – x)3

Conversely, we can have g ೦ f, where (g ೦ f)(x) = g (f(x)) So that f(g(x)) = g(x3) = 2 – x3

So now, you can see that, given two functions, you can substitute one into the other, and you'll wind up with an entirely different function. The process of composition also applies to three or more functions; you just keep substituting one function into the other, until you've completely substituted all given functions. For example, suppose you have three functions: f, g and h. Therefore, (f ೦ g ೦ h)(x) = f (g(h(x))) Note that when substituting, you do it in reverse order. In other words, start with h, then g and finally f.

Try the exercise below: If f(x) = 2 – cos x g(x) = 1/x2 h(x) = sin x, Find (I) f ೦ g

(II) g ೦ f

(III) f ೦ h (IV) h ೦ f

(V) g ೦ h

(VI) h ೦ g

(VII) f ೦ g ೦ h (VIII) h ೦ g ೦ f

Just as we use composition to combine functions, it can also be used to “decompose” a function into its component functions. Here's how it works: Suppose we have a function G(x) = (x – 1)3 + 2. How do we split G into its component functions? The key is understanding how G is built up: STEP 1: start with x

⇒

x

STEP 2: subtract 1 from x

⇒

x–1

STEP 3: find the cube of the result

⇒

x3

STEP 4: add 2 to the result

⇒

x+2

STEP 1 is always the basic first step; every function is built from the single variable. So now, we have three functions: x – 1, x3, and x + 2. Each function will be represented by three letters of your choice IN REVERSE ORDER. In this case, we use the letters f, g, and h. So, we have h(x) = x – 1 g(x) = x3 f(x) = x + 2 Since G has been split into three functions, you can check your answer by computing f ೦ g ೦ h and see if it equals G. Therefore, (f ೦ g ೦ h)(x) = f (g(h(x)))

= f (g(x – 1)) = f ((x – 1)3) = (x – 1)3 + 2

=

G(x)

I hope you understand how composition works. Try the following exercises, just to brush up: (1). If H(x) = 2 – cos x, express H in such a way that H = f(g(x)). In other words, split H into two functions, f and g, such that H = f ೦ g (2). Decompose each function into its component functions: a)

y = tan 2x

b) y = cos(x/2) c)

y = 1/(x+3)

d) 2 – √x + 1

Now, let's go back to the Continuity Theorems.

We continue with Theorem 3: Suppose we have a function g, which is continuous at the number a. Based on the definition of a limit, if x→a, then g(x)→g(a). Thus, we say that

lim g(x) = g(a)

x→a Now, let's assume that we have another function f, which just happens to be continuous at g(a). Thus, if g(x)→g(a), then

f(g(x))→f(g(a)). For simplicity, let's put g(a) = b, which leads to Theorem 3:

Theorem 3: If f is continuous at b and

lim g(x) = g(a) = b

x→a then

lim f(g(x))

=

lim f(g(x))

= f(b)

x→a

f

lim g(x)

x→a

OR x→a And then, we have Theorem 4:

Theorem 4: If g is continuous at a and f is continuous at g(a), then the composite function f ೦ g is continuous at a.

OR If h(x) = f(g(x)), then h is continuous at a.

Essentially, Theorem 4 simply states that “a continuous function of a continuous function is a continuous function.”

Next, we shall treat some examples, which will illustrate the use of the four Continuity Theorems. In the first few examples, we'll show why each function is continuous at every number in its domain, after which we'll determine the domain of the function.

Example 2 f(t) = 2t +

√25 – t2

Solution The first step to solving the problem is understanding the kind of function we're dealing with. In this case, we see that f can be expressed as a sum: f(t) = g(t) + h(t) where g(t) = 2t h(t) =

√25 – t2

By Theorem 2, g is continuous everywhere, since it is a polynomial. As a result, its domain is (– ∞,∞). As for h, it is continuous everywhere, provided that h(t) ≥ 0, that is, as long as t ≤ ± 5. Based on this, the domain of h is [–5, 5]. g and h have two different domains, and so we need to combine the domains.

When combining functions with different domains, the resulting domain is formed by the intersection of the individual domains. Since the domain of g is (–∞,∞), and the domain of h is [–5, 5], then the domain of g + h is (–∞,∞) ∩ [–5, 5]

=

[–5, 5]

So, by Part 1, f is continuous on the interval [–5, 5]

Just to be sure you are right (especially about the domain), you can graph the function. Below is a graph of f:

Example 3 h(x) =

5

√x – 1

(x2 – 2)

Solution Observe h carefully, and you'll see that it is a product: h(x) = (fg)(x), where f(x) =

5

√x – 1

g(x) = x2 – 2 ➔ f is defined and continuous everywhere, that is, on (–∞,∞). Observe that we are dealing with an odd root. Odd roots are defined for all values of x (Part II, Theorem 2) ➔ g is also continuous on (–∞,∞), since it is a polynomial (Parts 1 & 3, Theorem 2) Since both f and g are continuous everywhere, it follows by Part IV of Theorem 1 that the product fg is also continuous everywhere. Try graphing f, and you'll see that it is indeed continuous everywhere.

Example 4 h(x) =

sin x x+1

Solution It is quite obvious that h is a rational function, which means that it is continuous everywhere, except when the denominator, x + 1 equals zero, This, of course happens when x = –1. This implies that the domain of h is

{x|x ≠ –1} = (–∞, –1) ∪ (–1, ∞) h is graphed below:

Example 5 h(x) = cos (1 – x2)

Solution Take a good look at the function, and ask yourself: is there any way this function could have been formed by simple

combination?

Definitely NOT!! Instead, you'll see that h seems to be a “function within a function”. In other words, h is composed of two or more functions.

We need to find

the component functions of h. To do this, start by determining which part of the function can be

replaced by a single variable. In this case, we can easily say x = 1 – x2, which means h(x) = cos x. That's it!! So, we have two functions: f(x) = cos x, and g(x) = 1 – x2. You'll see that h(x) = f(g(x)).

Next, we examine the component functions: ➔ f(x) = cos x is continuous everywhere, since it is a trigonometric function. And so, its domain is (–∞, ∞), by Part III of Theorem 2. ➔ g(x) = 1 – x2 is polynomial, which means it is also continuous everywhere.

Thus, by Theorem 4, h is continuous on (–∞, ∞). h is graphed below:

Example 6 h(x) = tan 2x

Solution There are two ways proving that h is continuous: ➔ By expressing it as a composite function; h(x) = f(g(x)) where f(x) = tan x g(x) = 2x OR ➔ By expressing it as a quotient of two functions:

tan 2x =

sin 2x cos 2x

It will be relatively easier to use the second method. So, we'll go for it.

By Part 5 of Theorem 1, we know that

tan 2x =

sin 2x cos 2x

which means that y = tan 2x is continuous everywhere except when cos 2x = 0. This happens when x is an odd integer multiple of

π/4.

See the graph of y = cos 2x below:

From the graph, you can see that cos 2x= 0 when x = ± π/4, ± 3π/4, ± 5π/4, ± 7π/4, and so on.

Thus, the function h(x) = tan 2x has discontinuities at x = ± π/4, ± 3π/4, ± 5π/4, ± 7π/4, ± 9π/4, ± 11π/4, ± 13π/4, etc.

We can therefore state the domain of h as

{x ε R| cos 2x ≠ 0} OR

{x ε R| x ≠ ± πn/4} h is graphed below:

Example 7 F(x) =

√x

sin x

where n is an odd integer

Solution First, express F as a product: F(x) = (fg)(x) where f(x) =

√x

g(x) = sin x ➔ f is a root function, and is therefore defined for all non-negative values of x. In other words, its domain is [0, ∞) ➔ g is defined for all x, since it is a trigonometric function Therefore, the domain of F is obtained by the intersection of the domains of f and g: Domain of F =

[0, ∞)

∩ (–∞,∞)

= [0, ∞)

So, the domain of F is [0, ∞) .

F is graphed below:

Example 8 F(x) = sin (cos (sin x))

Solution Take a very good look at F, and you'll see that it is a composition of three functions: sin (cos (sin x))

=

f (g (h(x))

Which means f(x) = sin x, g(x) = cos x, h(x) = sin x

Since f, g and h are trigonometric functions, they are continuous everywhere (Part III, Theorem 2). This implies that the domain of F is (–∞,∞). F is graphed below:

Example 9 h(x) =

4

√x

+

x3 cos x

Solution Clearly, h can be expressed as as sum: h(x) = (f + g)(x), where f(x) =

4

√x

g(x) = x3 cos x ➔ f is continuous on [0, ∞) , by Part III, Theorem 2. ➔ g is a product of two continuous functions: x3 (a polynomial) and cos x (a trigonometric function). Both are continuous everywhere; thus the domain is (–∞,∞).

Thus, by Part I of Theorem 1, h is continuous on its domain, which is obtained from the intersection of the domains of f and g: [0, ∞)

∩ (–∞,∞)

= [0, ∞)

Hence, h is continuous on [0, ∞).

Example 10 h(x) =

√x2 – 9 x2 – 2

Solution h is a quotient function. Therefore, we can say f(x) =

√x

2

–9

2

g(x) = x – 2 We see that h is a rational function; which means that it is continuous everywhere, except when x 2 – 2 = 0. This occurs when x = √2. Hence, the domain of h, based on the denominator, is

{x ε R| x ≠ √2} 2

(Note that the function y = x – 2 itself is a polynomial, and is continuous everywhere) At

the same time, we need to consider the numerator, especially since it is not continuous everywhere. f

is

also

rational function, and is continuous on its domain, as long as f(x) ≥ 0. In other words, f is continuous for all x, provided that x ≤ –3 and x ≥ 3. Therefore, the domain of f is (–∞, –3]

∪ [3, ∞)

Therefore, Domain of h

∪ [3, ∞) ∩ {x ε R| x ≠ √2}

=

(–∞, –3]

=

(–∞, –3] ∪ [3, ∞)

as the graph of h clearly shows:

Example 11 Locate the discontinuities of the function and illustrate by graphing.

G(x) =

1 1 + sin x

Solution G is a rational function, which means that it is defined (and therefore continuous) for all x, except where 1 + sin x = 0. this happens when x = –π/2. This, of course isn't the only point of discontinuity. As the graph below illustrates:

As you can see from the graph, 1 + sin x = 0 when x = –13π/2, –9π/2, –5π/2, –π/2, 3π/2, 7π/2, 11π/2, and so on.

Obviously, there are many values to deal with. So, we'll need a general formula. Clearly, there seems to be some kind of pattern here; an arithmetic progression. In Sequences and Series, recall that the formula for the nth term of a sequence is given by

Tn = a + (n – 1)d Where Tn is the nth term, a is the first term n is the term to be found d is the common difference between the terms.

Here, the sequence is –13π/2, –9π/2, –5π/2, –π/2, 3π/2, 7π/2, 11π/2 The very first value of x for which 1 + sin x = 0 is –π/2. Thus a = –π/2. The common difference, d , is 2π.

One thing you need to understand about this sequence is that, although it is linear, it goes both ways. In other words, – π/2 may be the first term, but there are many others before. Thus, we need to able to find the remaining members of the sequence, regardless of their positions in the sequence. This means we remove (n – 1) in the formula and replace with n.

This gives

Tn = a + nd

So now, we have a general formula which enables us to locate the discontinuities of G. Using the formula above, we have

x = –π/2 + 2πn Thus, G has discontinuities at x = –π/2 + 2πn, where n is an integer. The graph below shows G:

When we were treating limit laws, we found that most limits can be evaluated by direct substitution. We also found that functions with this property are said to be continuous. So now we will use the concept of continuity to evaluate the limits, as the next two examples illustrate:

Example 12 Use continuity to evaluate the limit.

lim

x→4

5 + √x

√5 + x

Solution To use continuity to evaluate the limit, ➔ DETERMINE IF THE FUNCTION IS CONTINUOUS ON ITS DOMAIN

In this case let F(x) =

5 + √x

√5 + x

Observe that F is a rational function. Therefore, it is continuous on its domain (Parts II and III, Theorem 2). ➔ DETERMINE THE DOMAIN OF THE FUNCTION Since f is a rational function, then its domain must include all real numbers such that

√5 + x

≠ 0

which means F is discontinuous at x = –5. In other words, F is continuous for all x, except –5. Also, let's not forget that the numerator is a root function with a “limited domain”, which is [0, ∞). Therefore, the domain of F is

{x ε R| x ≠ – 5}

∩

[0, ∞)

=

[0, ∞)

➔ EVALUATE THE LIMIT VIA DIRECT SUBSTITUTION, IF THE NUMBER a BELONGS TO THE DOMAIN OF F: Since a = 4 belongs in the domain of F, which is [0, ∞), then we can evaluate the limit by substituting directly.

Therefore,

5 + √x

lim

x→4

=

√5 + x

F(4)

=

5 + √4

√5 + 4

=

=

5+2

√9 7 3

So,

lim

x→4

5 + √x

=

√5 + x

7 3

Example 13 Use continuity to evaluate the limit.

lim sin (x + sin x)

x→π

Solution Let h(x) = sin (x + sin x) Observe the h is a composite function of the form h(x) = f(g(x)), where f(x) = sin x g(x) = x + sin x Clearly, f and g are continuous everywhere, and this allows us to easily compute the value of the limit via direct substitution:

lim sin (x + sin x)

x→π

=

h(π )

=

sin (π + sin π )

=

sin (π + 0)

=

sin (π )

=

0

OR we could use the trigonometric addition formula. Recall that Sin (A + B) = Sin A Cos B + Cos A Sin B Therefore, Sin (π + sin π) = Sin π Cos(sin π) + Cos π Sin(Sin π) = 0 × Cos (0) + –1 × Sin (0) = (0 × 1) + (–1 × 0) =

0

Before we conclude this section, take a look at the following problems and solutions, after which you'll try some exercises.

Problem 14 Prove that f is continuous a if and only if

lim f (a + h) = f(a)

h→0

Solution To start with, let's graph f (we simply draw a simple, continuous curve):

h

I believe the graph above pretty much speaks for itself. Here, we see that the variable is h. It is also clear that as h decreases, (a+h) approaches a. So, if f is continuous, we find that, as (a+h)→ a, then f(a+h)→ f(a).

From the graph, let's assume that h keeps decreasing in value until it eventually becomes zero. Thus, we get to a point where (a+h) = a and so, f(a+h) = f(a).

Thus, from the definition of a limit and continuity,

lim f (a + h) = f(a)

h→0

Now here's a problem for you to solve: Using example 14 and the fact that

lim Sin θ = 0

θ→0

and

lim Cos θ = 1

θ→0

Show that

lim sin (a + h) = sin a

h→0

and

lim cos (a + h) = cos a

h→0

HINT: Apply the addition formulas: Sin (A + B) = Sin A Cos B + Cos A Sin B Cos (A + B) = Cos A Cos B – Sin A Sin B

Problem 15 For what values of x is continuous?

f(x) =

0

if x is rational

1

if x is irrational

Solution Before we can solve this problem, we need to re-examine the concept of rational and irrational numbers.

By definition, ➔ A rational number is a real number that can be expressed as a ratio of integers

NOTE: The decimal form of a rational number is repeating. For example, 1/2 = 0.5000000000 1/3 = 0.3333333333 2/3 = 0.6666666666 1/9 = 0.1111111111 3/11 = 0.2727272727 ➔ An irrational number is a real number that CANNOT be expressed as a ratio of integers

NOTE: The decimal form of a rational number is non-repeating. For example, √2 = 1.414213562

π

= 3.14159265

log104 = 0.3010299957 sin 130 = 0.4201670368

Now that we have confirmed what rational and irrational numbers are, we can easily construct a domain. So, ➔ f(x) = 0 if x is rational. Therefore, DOMAIN OF f = {x ε R| x = RATIONAL NUMBER} ➔ f(x) = 1 if x is irrational. Therefore, DOMAIN OF f = {x ε R| x = IRRATIONAL NUMBER} Clearly, both domains have absolutely nothing in common. Therefore, f is discontinuous for all x.

The following problem illustrates how we can apply the concept of continuity to physical phenomena.

Problem 15 The gravitational force exerted by Earth on a unit mass at a distance r from the center of the planet is

GMr R3

if r < R

GM r2

if r ≥ R

F(r) =

Where M is the mass of Earth, R is its radius and G is the gravitational constant. Is F a continuous function of r?

Solution This function presents four different values: three constants and one variable. The variable is r and the three constants are: ➔

The Gravitational Constant, G:

6.67300 × 10

➔

Mass of Earth, M:

5.9742 × 10

➔ Radius of Earth, R:

–11

24

m 3 kg

–1

s

–2

kg

6378.1 km

These values were obtained from Google. Alternatively, you can get approximate values from Wikipedia.

F is clearly a piecewise defined function, which can be simplified by inserting the values of the constants:

(6.67300 × 10 –11 × 5.9742 × 10 24)r (6378.1) 3

if r < R

F(r) = 6.67300 × 10 –11 × 5.9742 × 10 24 (6378.1) 2

if r ≥ R

Which gives

F(r) =

1536.479995r

if r < 6378.1

3.98658366 × 10 14 r2

if r ≥ 6378.1

This makes it a lot easier to graph F, so as to have a visualization of what we're dealing with:

The red curve represents F(r) = GMr/R3 (when r < Earth Radius) and the blue curve represents F(r) = GM/r2 (when r ≥ Earth Radius).

Now we need to determine if F is continuous. To do this, we have to show that F meets the conditions for continuity:

➔ F is clearly defined (and therefore continuous on its domain [0, ∞)). R belongs in this domain, and specifically, we see that

F(R) = 9.8 × 106 ➔ The left and right hand limits exist and are equal:

lim F (r) = F(R) = 9.8 × 106

r→R–

lim F (r) = F(R) = 9.8 × 106

r→R+

All conditions for continuity have been satisfied, and we can say that F is indeed a continuous function of r .

Note that physical phenomena are usually continuous.

You can read more about Earth Mass, Earth Radius, the Gravitational Constant, and Earth at Wikipedia.

Problem 16 Show that the absolute value function F(x) = |x| is continuous everywhere.

Solution By definition,

F(x) = |x| =

–x x

if x < 0 if x ≥ 0

The absolute value function is defined by two functions with two different domains: ➔

F(x) = –x if

x < 0. Therefore, its domain is (–∞, 0)

➔ F(x) = x if x ≥ 0. Therefore, its domain is [0, ∞)

F is composed of two functions y = x and y = – x, and so its domain will be formed by the union of the two domains.

Therefore, Domain of y = |x| = (–∞, 0) ∪ [0, ∞) = (–∞, ∞) Since the domain of F(x) = |x| is (–∞, ∞), it follows that this function is continuous everywhere, as its graph illustrates:

Exercises (1) Write an equation that expresses the fact that a function is continuous at the number 4 (2) If f is continuous on (–∞, ∞) , what can you say about its graph?

(3) Sketch a function that is continuous everywhere except at x = 3 and is continuous from the left at 3.

(4) Sketch the graph of a function that has a jump discontinuity at x = 2 and a removable discontinuity at x = 4 but is continuous elsewhere.

(5) A parking lot charges $3 for the first hour (or part of an hour) and $2 for each succeeding hour (or part), up to a daily maximum of $10 (I)

Sketch a graph of the cost of parking at this lot as a function of the time parked there.

(II)

Discuss the discontinuities of this function and their significance to someone who parks in the lot.

(6) Explain why each function is continuous or discontinuous (I)

The temperature at a specific location as a function of time

(II)

The temperature at a specific time as a function of the distance due west from New York City.

(III)

The altitude above sea level as a function of the distance due west from New York City.

(IV)

The cost of a taxi ride as a function of the distance traveled.

(V)

The current in the circuit for the lights in a room as a function of time.

(7) Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval:

F(x) =

x+1 x–3

(–∞, 3)

(8) Explain why the function is discontinuous at the given number. Sketch the graph of the function: (I)

(II)

f(x) =

f(x) =

x2 – 1 x+1

a = –1

x2 – 2x – 8 x–4

if x ≠ 4

3

if x = 4

a=4

(9) Locate the discontinuities of the function y = tan √x and illustrate by graphing.

(10) Prove that if f is a continuous function on an interval, then so is Also, if | f | is continuous, does it follow that f is continuous? If yes, prove it. If not, find a counterexample.

(11) For what values of x is g continuous?

g(x) =

0

if x is rational

x

if x is irrational

(12)Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x ≠ a and is continuous everywhere.

In

(I)

f(x) =

(II)

f(x) =

x2 – 2x – 8 x+2 x–7 |x – 7|

a = –2

(III)

f(x) =

x3 + 64 x+4

a = –4

a =7

(IV)

f(x) =

3 – √x 9–x

a =9

the next tutorial, we focus on a very important property of continuous functions, a property that is

expressed as what is known as the Intermediate Value Theorem.

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