On-line Math 21

1.5  Precise Definition of a Limit

Example 1 Show that

lim x® 1  3x+5 = 8.

Solution

In order to make 3x+5 be as close as we need to 8 , we just have to take x close enough to 1 . That is what we have to show. How close does x have to be to 1 ? How close do we need 3x+5 to be to 8 ? We presume that some unknown positive distance e has been chosen as the closeness we need to get 3x+5 to 8 . That is, we need to arrange for

|(3x+5)-8| < e.

What are we allowed to do to arrange this? We can only make x closer to 1 - as close as we need to 1 . We can control that distance. The key to figuring these out is to deal with the expression we need to control (the |(3x+5)-8| in this case), and hope that within that expression we can see the quantity we can control (the |x-1| ). But, this time,

|(3x+5)-8| = |3x-3| = 3|x-1|.

so we can control the distance from 3x+5 to 8 just by making x close to 1. No matter what e we have, we can get 3x+5 within e of 8 just by taking |x-1| < e/3 . That is, we take d = e/3 , and whenever ) < |x-1| < d, we will have (3x+5)-8| < e.

Why divide by 3? We do that to cover for the 3 in front of the |x-1| .

So, what is the answer? The closest thing to an answer here is the choice of d depending on e, d = e/3 . But there is more to be said than just that. You have to justify your ``answer'', and explain why it works. That is, you need to prove that the limit is as you say it is.

The proof you should write

Let e > 0 be given. If d = e/3 , then whenever

0 < |x-1| < d,

|(3x+5)-8| = 3|x-1| < 3 æ ç è e 3 ö ÷ ø = e.

So,

lim x® 1  3x+5 = 8.

Notice that the string of equations/inequalities, following from beginning to end, does say that |(3x+5)-8| < e. It is typical to write an equals sign in an intermediate step if it really is equal to the next step. The total relation may end up as an inequality, as it did here, though. Note also that I used the calculations I did earlier in my ``scratch work'' to get from the |f(x)-L| back to the |x-a| .

Copyright (c) 2000 by David L. Johnson.