DEFINITION OF CONTINUITY OF A FUNCTION AT A POINT. A function is said to be continuous a point if
(a) is defined at , and
is continuous at
if for every
there is a
EXAMPLE 1. Constant functions are continuous everywhere. If
for every , so is continuous everywhere.
EXAMPLE 2. The identity function is continuous everywhere. If
for every , so the identity function is continuous everywhere.
EXAMPLE 3. Let for all This function is continuous at every point which is not an integer. At the integers it is discontinuous, since the limit of does not exist, the right and left-hand limits being unequal. A discontinuity of this type, where the right- and left-hand limits exist but are unequal, is called a jump discontinuity.
EXAMPLE 4. The function for which for ,, is discontinuous at . We say there is an infinite discontinuity at because the function takes arbitrarily large values near
EXAMPLE 5. Let for . This function is continuous everywhere except at . It is discontinuous at because is not equal to the limit of as . In this example, the discontinuity could be removed by redefining the function at to have the value instead of . For this reason, a discontinuity of this type is called a removable discontinuity. Note that jump discontinuities, such as those possessed by the greatest-integer function, cannot be removed by simply changing the value of fat one point.
Calculations with limits may often be simplified by the use of the following theorem which provides basic rules for operating with limits.
be functions such that
Then we have
Proofs of (i) and (ii). Since the two statements
are equivalent, and since we have
it suffices to prove part (i) of the theorem when the limits and are both zero. Suppose, then, that and as . We shall prove that as . This means we must show that for every there is a such that
Let be given. Since as , there is a such that
Similarly, as , there is a such that
If we let denote the smaller of the two numbers and , then the above two inequalities are valid if and hence, by the triangle inequality, we find that
This proves (i). The proof of (ii) is entirely similar, except that the last step we use the inequality
Proof of (iii). Suppose that we have proved part (iii) for the special case in
which one the limits is
Then the general case follows easily from this special case. In fact, all we
need to do is write
The special case implies that each term on the right approaches as and, by property (i), the sum of the two terms also approaches . Therefore, it remains to prove (iii) in the special case where one of the limits, say , is .
Suppose, then, that
We wish to prove that
To do this we must show that if a positive
is given, there is a
Since as , there is a such that
For such ,. we have
Since as , for every there is a such that
Therefore, if we let be the smaller of the two numbers and , then both inequalities are valid whenever , and for such we deduce the desired inequality
This completes the proof of (iii).
Proof of (iv). Since the quotient
is the product of
it suffices to prove that
and then appeal to (iii). Let
and we wish to prove that
be given. We must show that there is a
The difference to be estimated may be written as follows.
Since as , we can choose a such that both inequalities
are satisfied whenever . The second of these inequalities implies so for such x. Using this in
along with the inequality , we obtain
This completes the proof of (iv).