| f(x) = | 2x-1 | if | x 3 |
| 6 | if | x = 3 |
Intuitively it is clear that when x is close to 3 but x
3,
then f(x) is close to 5, and so lim x --> 3 f(x) = 5.
To obtain more detailed information about how f(x) varies when x is close to 3, we ask the following question:
The distance from x to 3 is | x - 3 | and the distance from f (x) to 5 is | f (x) -5 |,
so our problem is to find a number
such that
but x
3
3, so an equivalent formulation of our problem is to find a number
such that
Notice that if 0 < | x - 3 | < (0.1)/2 = 0.05, then
Thus, an answer to the problem is given by
= 0.05; that is, if x is within a distance of 0.05 from 3,
then f (x) will be within a distance of 0.1 from 5.
If we change the number 0.1 in our problem to the smaller number 0.01, then by using the same method we find that f (x) will differ from 5 by less than 0.01 provided that
x differs from 3 by less than (0.01)/2 = 0.0005:
(the Greek letter epsilon),
then we find as before that
if 0 < | x - 3 | <
=
/2
from 5 by taking the values of x within a distance
/2 from 3 (but x
3).
< f (x) < 5 +
whenever 3 -
< x < 3 +
(x
3)
3) to like in the interval (3 -
, 3 +
) we can make the values of f (x) lie in the interval (5 -
.
Using (1) as a model, we give a precise (formal) definition of a limit,

there is a corresponding number
such that
whenever
|
|