Calculus of One and Several Variables

Chapter 4
Continuous Functions and Limits of Functions

In the last chapter, the idea of a function was described and the principal ideas related to functions of natural numbers (sequences) were presented. However, the concept of function is far too general to be useful in calculus. There are various ways to restrict the concept in order to study something interesting and the types of restrictions considered depend very much on what you find interesting. In Calculus, the most fundamental restriction made is to assume the functions are continuous. Continuous functions are those in which a sufficiently small change in x results in a small change in f

. They rule out things which could never happen physically. For example, it is not possible for a car to jump from one point to another instantly. Making this restriction precise turns out to be surprisingly difficult although many of the most important theorems about continuous functions seem intuitively clear.

Before giving the careful mathematical definitions, here are examples of graphs of functions which are not continuous at the point x₀.

You see, there is a hole in the picture of the graph of this function and instead of filling in the hole with the appropriate value, f

is too large. This is called a removable discontinuity because the problem can be fixed by redefining the function at the point x₀. Here is another example.

You see from this picture that there is no way to get rid of the jump in the graph of this function by simply redefining the value of the function at x₀. That is why it is called a nonremovable discontinuity or jump discontinuity. Now that pictures have been given of what it is desired to eliminate, it is time to give the precise definition.

The definition which follows, due to Cauchy,¹ Bolzano, and Weierstrass² is the precise way to exclude the sort of behavior described above and all statements about continuous functions must ultimately rest on this definition from now on.

In sloppy English this definition says roughly the following: A function f is continuous at x when it is possible to make f

as close as desired to f provided y is taken close enough to x. In fact this statement in words is pretty much the way Cauchy and Bolzano described it. The completely rigorous definition above is due to Weierstrass. This definition does indeed rule out the sorts of graphs drawn above. Consider the second nonremovable discontinuity. The removable discontinuity case is similar.

For the ε shown, you can see from the picture that no matter how small you take δ, there will be points x, between x₀ −δ and x₀ where f

< 2 −ε. In particular, for these values of x, > ε.  Therefore, the definition of continuity given above excludes the situation in which there is a jump in the function. Similar reasoning shows it excludes the removable discontinuity case as well. There are many ways a function can fail to be continuous and it is impossible to list them all by drawing pictures. This is why it is so important to use the definition. The other thing to notice is that the concept of continuity as described in the definition is a point property. That is to say it is a property which a function may or may not have at a single point. Here is an example.

To verify the assertion about the above function, first show it is not continuous at x if x≠0. Take such an x and let ε =

∕2. Now let δ > 0 be completely arbitrary. In the interval, there are rational numbers, y₁ such that > and irrational numbers, y₂. Thus = > . If f were continuous at x, there would exist δ > 0 such that for every point, y ∈, < ε. But then, letting y₁ and y₂ be as just described,

|x|	< |y1| = |f (y1) − f (y2)|
	≤ |f (y1)− f (x)| + |f (x)− f (y2)| < 2ε = |x| ,

which is a contradiction. Since a contradiction is obtained by assuming that f is continuous at x, it must be concluded that f is not continuous there. To see f is continuous at 0, let ε > 0 be given and let δ = ε. Then if

< δ = ε, Then

|f (y)− f (0)|	= 0 if y is irrational
|f (y)− f (0)|	= |y| < ε if y is rational.

either way, whenever

< δ, it follows < ε and so f is continuous at x = 0. How did I know to let δ = ε? That is a very good question. The choice of δ for a particular ε is usually arrived at by using intuition, the actual εδ argument reduces to a verification that the intuition was correct. Here is another example.

To do this, note first that f

= 25 and it is desired to verify the conditions for continuity. Consider the following.

|− 5x + 10− (25)| = 5 |x − (− 3)|.

This allows one to find a suitable δ. If ε > 0 is given, let 0 < δ ≤

ε. Then if 0 < < δ, it follows from this inequality that

1 |− 5x+ 10 − (25)| = 5|x− (− 3)| < 55 ε = ε.

Sometimes the determination of δ in the verification of continuity can be a little more involved. Here is another example.

First note f

= . Now consider: 

||√ ------- √ --|| || 2x + 12− 22 || | 2x+ 12 − 22| = ||√--------√---|| 2x + 12+ 22

2 1 √-- = √2x-+-12-+√22-|x − 5| ≤ 11 22 |x − 5|

whenever

< 1 because for such x, > 0. Now let ε > 0 be given. Choose δ such that 0 < δ ≤ min. Then if < δ, all the inequalities above hold and

||√------- √--|| 2 2 ε√22 | 2x + 12− 22| ≤ √22-|x − 5| < √22--2--= ε.

First observe f

= −140. Now

|| 2 || − 3x + 7− (− 140) = 3|x+ 7||x− 7| ≤ 3(|x|+ 7)|x − 7|.

If

< 1, it follows from the version of the triangle inequality which states ≤ that < 1 + 7. Therefore, if < 1, it follows that

| | |− 3x2 + 7− (− 140)| ≤ 3((1+ 7)+ 7)|x− 7|

= 3(1+ 27)|x− 7| = 84|x− 7|.

Now let ε > 0 be given. Choose δ such that 0 < δ ≤ min

. Then for < δ, it follows

| | ( ε) |− 3x2 + 7 − (− 140)| ≤ 84 |x − 7| < 84 84 = ε.

The following is a useful theorem which can remove the need to constantly use the ε,δ definition given above.

Proof: First consider 1.) Let ε > 0 be given. By assumption, there exist δ₁ > 0 such that whenever

< δ₁, it follows < and there exists δ₂ > 0 such that whenever < δ₂, it follows that < . Then let 0 < δ ≤ min. If < δ, then everything happens at once. Therefore, using the triangle inequality

|af (x) + bf (x)− (ag (y) + bg (y))|

	≤ |a| |f (x)− f (y)| + |b| |g(x)− g (y)|
	< |a| ( ) -----ε------- 2(|a|+ |b|+ 1) + |b| ( ) ------ε------ 2 (|a|+ |b|+ 1) < ε.

Now consider 2.) There exists δ₁ > 0 such that if

|y − x| < δ , 1

then

< 1. Therefore, for such y, < 1 + . It follows that for such y,

|fg(x)− fg (y)| ≤ |f (x)g(x)− g (x)f (y)|+ |g(x)f (y)− f (y)g (y)|

	≤ |g(x)| |f (x) − f (y)| + |f (y)| |g(x)− g(y)|
	≤ (1+ |g(x)|+ |f (y)|) [|g(x)− g(y)|+ |f (x)− f (y)|]
	≤ (2+ |g(x)|+ |f (x)|) [|g(x)− g(y)|+ |f (x) − f (y)|]

Now let ε > 0 be given. There exists δ₂ such that if

< δ₂, then

ε |g(x)− g (y)| <-------------------, 2 (1 + |g (x)|+ |f (y)|)

and there exists δ₃ such that if

< δ₃, then

---------ε--------- |f (x)− f (y)| < 2(2+ |g(x)|+|f (x)|)

Now let 0 < δ ≤ min

. Then if < δ, all the above hold at once and so

|fg(x)− fg (y)| ≤

	(2 + |g (x)|+ |f (x)|) [|g (x)− g(y)|+ |f (x)− f (y)|]
	< (2+ |g(x)|+ |f (x)|) ( ) --------ε---------- --------ε---------- 2(2+ |g (x )|+ |f (x)|) + 2(2+ |g(x)|+ |f (x)|) = ε.

This proves the first part of 2.) To obtain the second part, let δ₁ be as described above and let δ₀ > 0 be such that for

< δ₀,

|g (x) − g(y)| < |g(x)|∕2

and so by the triangle inequality,

− |g(x)|∕2 ≤ |g(y)|− |g(x)| ≤ |g(x)|∕2

which implies

≥∕2, and < 3∕2.

Then if

< min,

||f (x) f (y)|| ||g(x) −-g(y)||	= ||f (x )g(y)− f (y)g(x)|| ||------g(x)g(y)-----||
	≤ |f (x)g((y)−-f2(y))-g(x)| |g(x2)|-
	= 2 |f (x)g (y) − f (y)g(x)| -------------2------- |g (x)|

	≤ 2 -----2 |g (x)| [|f (x)g (y) − f (y)g(y)+ f (y)g(y)− f (y) g(x)|]
	≤ --2--- |g (x)|2 [|g(y)||f (x )− f (y)|+ |f (y)||g (y)− g (x)|]
	≤ 2 -----2 |g (x)| [3 ] 2 |g (x)||f (x)− f (y)|+ (1 + |f (x )|)|g (y) − g (x)|
	≤ --2--2 |g (x)| (1+ 2|f (x)|+ 2|g(x)|) [|f (x)− f (y)|+ |g(y)− g(x)|]
	≡ M [|f (x)− f (y)|+ |g(y)− g(x)|]

where M is defined by

M ≡ ---2-- (1+ 2|f (x)|+2 |g (x)|) |g(x)|2

Now let δ₂ be such that if

< δ₂, then

ε −1 |f (x)− f (y)| < 2M

and let δ₃ be such that if

< δ₃, then

|g(y)− g(x)| < εM − 1. 2

Then if 0 < δ ≤ min

, and < δ, everything holds and

||f (x) f (y)|| ||-----− ----|| ≤ M [|f (x)− f (y)|+ |g(y)− g(x)|] g (x ) g (y)

[ε −1 ε −1] < M 2M + 2M = ε.

This completes the proof of the second part of 2.)

Note that in these proofs no effort is made to find some sort of “best” δ. The problem is one which has a yes or a no answer. Either it is or it is not continuous.

Now consider 3.). If f is continuous at x, f

∈ D ⊆ ℝ, and g is continuous at f,then g ∘f is continuous at x. Let ε > 0 be given. Then there exists η > 0 such that if