James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 2.4 The Precise Definition of a Limit 111
2. Showing that this works. Given « . 0, let 5 minh1, «y7j. If
, 7 (as in part l).
, «y7, so
0 , | x 2 3 | , , then | x 2 3 | , 1 ? 2 , x , 4 ? | x 1 3 |
We also have | x 2 3 |
| x 2 2 9 | − | x 1 3 | | x 2 3 | , 7 ? « 7 − «
This shows that lim x l3 x 2 5 9.
■
As Example 4 shows, it is not always easy to prove that limit statements are true
using the «, definition. In fact, if we had been given a more complicated function such
as f sxd 5 s6x 2 2 8x 1 9dys2x 2 2 1d, a proof would require a great deal of ingenuity.
Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be
proved using Definition 2, and then the limits of complicated functions can be found
rigorously from the Limit Laws without resorting to the definition directly.
For instance, we prove the Sum Law: If lim x l a f sxd 5 L and lim x l a tsxd 5 M both
exist, then
lim f f sxd 1 tsxdg − L 1 M
x l a
The remaining laws are proved in the exercises and in Appendix F.
Proof of the Sum Law Let « . 0 be given. We must find . 0 such that
if 0 , | x 2 a | , then | f sxd 1 tsxd 2 sL 1 Md | , «
Triangle Inequality:
(See Appendix A).
| a 1 b | < | a | 1 | b |
Using the Triangle Inequality we can write
5 | f sxd 1 tsxd 2 sL 1 Md | 5 | s f sxd 2 Ld 1 stsxd 2 Md |
< | f sxd 2 L | 1 | tsxd 2 M |
We make | f sxd 1 tsxd 2 sL 1 Md | less than « by making each of the terms | f sxd 2 L |
and | tsxd 2 M | less than «y2.
Since «y2 . 0 and lim x l a f sxd 5 L, there exists a number 1 . 0 such that
if 0 , | x 2 a | , 1 then | f sxd 2 L | , « 2
Similarly, since lim x l a tsxd − M, there exists a number 2 . 0 such that
if 0 , | x 2 a | , 2 then | tsxd 2 M | , « 2
Let − minh 1 , 2 j, the smaller of the numbers 1 and 2 . Notice that
if 0 , | x 2 a | , then 0 , | x 2 a | , 1 and 0 , | x 2 a | , 2
and so | f sxd 2 L | , « and
2
| tsxd 2 M | , « 2
Therefore, by (5),
| f sxd 1 tsxd 2 sL 1 Md | < | f sxd 2 L | 1 | tsxd 2 M |
, « 2 1 « 2 5 «
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