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

appendix F Proofs of Theorems A41
This shows that
and so, for these values of x,
if 0 , | x 2 a | , 1 then | tsxd | . | M |
2
Also, there exists 2 . 0 such that
1
| Mtsxd | − 1
| M | | tsxd | , 1
| M | ? 2
| M | − 2 M 2
if 0 , | x 2 a | , 2 then | tsxd 2 M | , M 2
Let − minh 1 , 2 j. Then, for 0 , | x 2 a |
, , we have
1
Z
tsxd 2 1 M
Z | − M 2 tsxd |
| Mtsxd , 2 M 2
| M 2 2 « − «
It follows that lim x l a 1ytsxd − 1yM. Finally, using Law 4, we obtain
2 «
lim
x l a
f sxd
tsxd − lim
x l aSf sxd ?
1
1
− lim f sxd lim
tsxdD x l a x l a tsxd − L ? 1 M − L M
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2 Theorem If f sxd < tsxd for all x in an open interval that contains a (except
possibly at a) and
then L < M.
lim f sxd − L and lim tsxd − M
x l a x l a
ProoF We use the method of proof by contradiction. Suppose, if possible, that L . M.
Law 2 of limits says that
lim ftsxd 2 f sxdg − M 2 L
x l a
Therefore, for any « . 0, there exists . 0 such that
if 0 , | x 2 a | , then | ftsxd 2 f sxdg 2 sM 2 Ld | , «
In particular, taking « − L 2 M (noting that L 2 M . 0 by hypothesis), we have a
number . 0 such that
if 0 , | x 2 a | , then | ftsxd 2 f sxdg 2 sM 2 Ld | , L 2 M
Since b < | b | for any number b, we have
if 0 , | x 2 a | , then ftsxd 2 f sxdg 2 sM 2 Ld , L 2 M
which simplifies to
if 0 , | x 2 a |
, then tsxd , f sxd
But this contradicts f sxd < tsxd. Thus the inequality L . M must be false. Therefore
L < M.
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