Consider the function f(n)=8n+128 shown in Figure .
Clearly, f(n) is non-negative for all integers
.
We wish to show that
.
According to Definition
,
in order to show this we need to find an integer
and a constant c>0
such that for all integers
,
.
It does not matter what the particular constants are--as long as they exist! E.g., suppose we choose c=1. Then
Since (n+8)>0 for all values of ,
we conclude that
.
I.e.,
.
So, we have that for c=1 and ,
for all integers
.
Hence,
.
Figure
clearly shows
that the function
is greater than
the function f(n)=8n+128 to the right of n=16.
Of course, there are many other values of c and that will do.
For example, c=2
and
will do,
as will c=4 and
.
(See Figure
).