On-line Math 21

1.5  Precise Definition of a Limit

Example 3 Show that

lim x® 3  1 x = 1 3 .

Hints

Again, you need to do a ballpark estimate. But, first you want to look at what you are trying to control, and re-express it to make |x-3| appear.

Next step

ê ê ê 1 x - 1 3 ê ê ê = |3-x| |3x| .

And then?

ê ê ê 1 x - 1 3 ê ê ê = |3-x| |3x| = 1 3|x| |x-3|.

We need to control the 1/3|x| term, since the other one we have complete control over.

Finish it

Again taking a ballpark of d = 1 , |x-3| < 1, and so

-1 < x-3 < 1,

or, adding 3 to all three sides,

2 < x < 4.

Thus, x > 2 . Flipping the terms reverses the inequality,

1 4 < 1 x < 1 2 ,

or (dividing again by 3),

1 12 < 1 3|x| < 1 6 ,

which means that all we have to do is cover a 1/6 to get within the needed bounds.

Here is the formal proof, which is what you should write down.

Let e > 0 be chosen. Then, set

d = min {1,6e}.

Whenever

0 < |x-3| < d

ê ê ê 1 x - 1 3 ê ê ê = 1 3|x| |x-3| £ 1 6 |x-3| < 1 6 (6e) = e.

Copyright (c) 2000 by David L. Johnson.