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Real Analysis
Notes where we have taken y = g(x). Hence h is continuous at a which proves the theorem.
Let us now study the following example:
Example: Examine for continuity the following functions:
(i) The polynomial function f R ® R defined by
n
f(x) = a, + a x + a x +... + a x .
2
1 2 x
(ii) The rational function f: R ® R defined as
f(x)
f(x) = , " x for which q(x) 0.
q(x)
Solution:
(i) It is obvious that the function f(x) = x, x R, is continuous on the whole of the real line. It
3
follows from theorem 3 that the functions x , x ,...., are all continuous. The fact that constant
Z
functions are continuous, we get that any polynomial f(x) in x, i.e., the function f defined
by
2
n
f(x) = a, + a x+ a x +... + a x ,
1 2 n
is continuous on R.
(ii) It follows from theorem 3(v) that a rational function f, defined by,
p(x) a + a x + + a x n
f(x) = = 0 1 n
q(x) b + b x + + b x m
m
0
1
is continuous at every point a R for which q(a) 0.
11.3 Non-continuous Functions
You have seen that a function may or may not be continuous at a point of the domain of the
function. Let us now examine why a function fails to be continuous.
A function f: S ® R fails to be continuous on its domain S if it is not continuous at a particular
point of S. This means that there exists a point a S such that, either
(i) lim f(x) does not exist, or
-
x a
(ii) lim f(x) exists but is not equal to f(a).
x a
-
But you know that a function f is continuous at a point a if and only if
f(a+) = f(a–) = f(a).
Thus, if f is not continuous at a, then one of the following will happen:
(i) either f(a+) or f(a–) does not exist (this includes the case when both f(a+) and f(a–) do not
exist).
(ii) both f(a+) and f(a–) exist but f(a+) f(a–).
(iii) both f(a+) and f(a–) exist and f(a+) = f(a–) but they are not equal to f(a).
If a function f: S ® R is discontinuous for each b S, then we say that totally discontinuous an S.
Functions which are totally discontinuous are often encountered but by no means rare. We give
an example.
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