On-line Math 21

1.5  Precise Definition of a Limit

Example 2 Show that

lim x® 2  x2+3x = 10.

Solution

Here there is an additional trick. We begin as we did before, looking at

|(x2+3x)-10|

and trying to show how that relates to |x-2| , which, in this example, is what we can control. Now,

|(x2+3x)-10| = |(x-2)(x+5)| = |x-2||x+5|.

So, in order to control |(x²+3x)-10| , we need also to get some control over |x+5| . Fortunately, we don't need all that much control over that term, we just need to be sure that it does not get too large.

The best approach to controlling that extra term is to find bounds on it in a set neighborhood of 2 . We need a ``ballpark estimate'' on |x+5| in some reasonable region (the ballpark). It is usually, but not always, safe to deal with the region one unit on either side of 2 , |x-2| < 1 . This region is also expressed as

-1 < x-2 < 1,

by unwinding the absolute value. Adding 2 to all three sides gives

1 < x < 3,

which solves for the x 's we are dealing with, and adding 5 more gives us bounds on the expression we are worried about,

6 < x+5 < 8.

Now, since x+5 is always positive, we don't really need to worry about the absolute value sign, but we'll keep it around. For all x in this ballpark, then,

|x+5| < 8,

so we only need to cover as much as 8 to get |(x²+3x)-10| small; that and the |x-2| over which we have direct control.

Notice that not only do we need d £ e/8 for the usual reason, but we also need to be sure that d £ 1 also. Of course, for a ``fair'' e this will not be a problem, but (in keeping with the baseball analogy) if someone throws us a curveball, we have to be ready for it. So, you have to stipulate that both d £ 1 and that d £ e/8 .

We're now ready for the formal proof:

Let e > 0 be chosen. Then, set

d =

min

{1,e/8}

. Whenever 0 < |x-2| < d, then

|(x2+3x)-10| = |x+5||x-2| £ 8|x-2|, since x is in the ballpark, < 8(e/8), the other choice of d, = e.

Thus

lim x® 2  (x2+3x) = 10.

Copyright (c) 2000 by David L. Johnson.