# Continuity of a Function

DEFINITION OF CONTINUITY OF A FUNCTION AT A POINT. A function is said to be continuous a point if

(a) is defined at , and

(b)

A function is continuous at if for every there is a such that

EXAMPLE 1. Constant functions are continuous everywhere. If for all , then

for every , so is continuous everywhere.

EXAMPLE 2. The identity function is continuous everywhere. If for all we have

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.

# The Basic Limit Theorems

Calculations with limits may often be simplified by the use of the following theorem which provides basic rules for operating with limits.

Theorem. Let and be functions such that

Then we have

(i)

(ii)

(iii)

(iv) if

# Proofs of the Basic Limit Theorem

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 and as . We wish to prove that as To do this we must show that if a positive is given, there is a such that

Since as , there is a such that

For such ,. we have

and hence

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 with , it suffices to prove that as and then appeal to (iii). Let . Then as , and we wish to prove that as . Let be given. We must show that there is a such that

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).

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