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
(x)
. 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 x_{0}.
PICT
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
(x0)
is too large. This is called a
removable discontinuity because the problem can be fixed by redefining the function at
the point x_{0}. Here is another example.
PICT
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 x_{0}. 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 Weierstrass^{2}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.
Definition 4.0.1A function f : D
(f)
⊆ ℝ → ℝ is continuous at x ∈ D
(f)
iffor each ε > 0 there exists δ > 0 such that whenever y ∈ D
(f)
and
|y − x| < δ
it follows that
|f (x) − f (y)| < ε.
A function f is continuous if it is continuous at every point of D
(f)
.
In sloppy English this definition says roughly the following: A function f is continuous
at x when it is possible to make f
(y)
as close as desired to f
(x)
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.
PICT
For the ε shown, you can see from the picture that no matter how small you take δ,
there will be points x, between x_{0}−δ and x_{0} where f
(x)
< 2 −ε. In particular, for these
values of x,
|f (x) − f (x0)|
> 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.
Example 4.0.2Let
{ x if x is rational
f (x ) = 0 if x is irrational .
This function is continuous at x = 0 and nowhere else.
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 =
|x|
∕2. Now let δ > 0 be completely
arbitrary. In the interval,
(x − δ,x + δ)
there are rational numbers, y_{1} such
that
|y |
1
>
|x|
and irrational numbers, y_{2}. Thus
|f (y) − f (y)|
1 2
=
|y |
1
>
|x|
.
If f were continuous at x, there would exist δ > 0 such that for every point,
y ∈
(x− δ,x+ δ)
,
|f (y)− f (x)|
< ε. But then, letting y_{1} and y_{2} 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
|y− 0|
< 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
|y− 0|
< δ, it follows
|f (y)− f (0)|
< 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.
Example 4.0.3Show the function f
(x)
= −5x + 10 is continuous at x = −3.
To do this, note first that f
(− 3)
= 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 ≤
1
5
ε. Then if
0 <
|x− (− 3)|
< 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.
This proves the first part of 2.) To obtain the second part, let δ_{1} be as described
above and let δ_{0}> 0 be such that for
|x− y|
< δ_{0},
|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
|g(y)|
≥
|g(x)|
∕2, and
|g(y)|
< 3
|g(x)|
∕2.
Then if
|x− y|
< min
(δ0,δ1)
,
||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 δ_{2} be such that if
|x− y|
< δ_{2}, then
ε −1
|f (x)− f (y)| < 2M
and let δ_{3} be such that if
|x− y|
< δ_{3}, then
|g(y)− g(x)| < εM − 1.
2
Then if 0 < δ ≤ min
(δ0,δ1,δ2,δ3)
, and
|x− y|
< 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
(x)
∈ D
(g)
⊆ ℝ, and g is continuous at
f
(x )
,then g ∘f is continuous at x. Let ε > 0 be given. Then there exists η > 0 such that
if