Page 157 - DMTH401_REAL ANALYSIS

Unit 11: Continuity

Notes
Example: Discuss the continuity of the function sin x on the real line R.

Solution: Let f(x) = Sin x " x  R.

We show by the (, )-definition that f is continuous at every point of R.

Consider an arbitrary point a  R. We have
x - a x + a
|f(x) – f(a)| = |sin x – sin a| = 2 sin cos
2 2

x - a x + a
= 2 sin cos
2 2
x - a æ x + a ö
 2 sin ç since cos  I ÷
2 è 2 ø
From Trigonometry, you know that |Sin |  ||.

x - a x - a x - a
Therefore sin  =
2 2 2
Consequently |f(x) – f(a)|  |x – a|

<  if |x – a] <  where  = .
So f is continuous at the point a. But a is any point of R. Hence Sin x is continuous on the real
line R.

Task Discuss the continuity cos x on the real R.

As we have connected the limit of a function with the limit of a sequence of real numbers. In the
same way, we can discuss the continuity of a function in the language of the sequence of real
numbers in the domain of the function. This is explained in the following theorem.
Theorem 2: A function f: S ® R is continuous at point a in S if and only for every sequence (x ),
n
(x  S) converging do a, f(x ) converges to f(a).
n n
Proof: Let us suppose that f is continuous at a. Then lim f(x) = f(a).
x® a
Given  > 0, there exists a  > 0 such that
|x – a| < 6  |f(x) – f(a)| < .

If x is a sequence converging to ‘a’, then corresponding to  > 0, there exists a positive integer M
n
such that
|x – a| <  for n  M.
n
Thus, for n  M, we have |x – a| <  which, in turn, implies that
n
|f(x ) – f(a)| < ,
n
proving thereby f(x ) converges to f(a).
n

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