**Theorem: **

If are continuous functions and c is a constant, then

a) is a continuous functions.

b) is a continuous functions.

c) cf is a continuous function.

d) fg is a continuous function.

**Proof:**

We shall only prove statement d. Choose and fix any . Since is continuous at , we have that for the positive number there exists a such that

whenever

Since , we conclude that

, whenever . Let . Also, since g is continuous at , for the positive number , there is a such that whenever . Put . Then, whenever , we have

equals

which equals

which equals

Observe that we have not claimed that the quotient of two continuous functions is continuous. The problem is obvious: cannot have any meaning at x for which . So, the question would be, if for every , is the function , defined by , continuous? The answer is yes. For a proof, we need a preliminary result.

**Lemma.**

if is continuous and , then there is an and such that if , then .

**Proof.**

Let . Now, . By continuity of g, there is a such that

for

But, and hence, $-m<|g(x)|-|g(x_{0})|<m$, giving us

for . Hence, the proof.

The lemma says that if a continuous function does not vanish at a point, then there is an interval containing it in which it does not vanish at any point.

**Theorem.**

If are continuous and for all x, then defined by is continuous.

*The proof of the above theorem using the lemma above is left as an exercise.*

**Examples.**

a) defined by for all , where is continuous.

b) defined by is continuous.

c) defined by is a continuous function because , where . Since f is continuous by (b), g must be continuous.

d) by , n being a positive integer, is continuous by repeated application of the above reasoning.

e) defined by , where is also continuous. This is because of the fact that if

are defined by , , …, , then are also continuous functions. Hence,

is also a continuous function as the sum of continuous functions is a continuous function. Thus, we have shown that a polynomial is a continuous function.

f) Let p and q be polynomials. Let be such that and for .

Now, let .

Then, defined by is a continuous function. What we have said is that a rational function which is defined everywhere except on the finite set of zeroes of the denominator is continuous.

g) defined by is continuous everywhere. Indeed, . Therefore,

(because , where x is measured in radians)

h) defined by is continuous since

i) defined by is continuous. We had to omit numbers like from the domain of f as cannot be defined for these values of x.

j) defined by is a continuous function. Indeed,

which equals

Choose to start with, so that . Thus,

Given an , let

.

Then, for , we have

.

It can be shown that f defined by is also a continuous function for every real .

k) Consider the function defined by . Is f a continuous function? *This is left as an exercise.* (**Hint**: It will suffice to prove continuity at . This would follow from ).

k) Suppose is defined by , if and . We can see that f is not continuous at 0 as changes abruptly when x goes over from negative to positive values.

More later,

Nalin Pithwa