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

A40
appendix F Proofs of Theorems
Since lim x l a f sxd − L, there is a number 3 . 0 such that
if 0 , | x 2 a | , 3 then | f sxd 2 L | , «
2(1 1 | M |)
Let − minh 1 , 2 , 3 j. If 0 , | x 2 a | , , then we have 0 , | x 2 a | , 1,
0 , | x 2 a | , 2, and 0 , | x 2 a | , 3, so we can combine the inequalities to
obtain
| f sxdtsxd 2 LM | < | f sxd 2 L | | tsxd | 1 | L | | tsxd 2 M |
«
,
2(1 1 | M |) (1 1 | M |) 1 | L «
|
2(1 1 | L |)
, « 2 1 « 2 − «
This shows that lim x l a f f sxd tsxdg − LM.
■
Proof of Law 3 If we take tsxd − c in Law 4, we get
lim fcf sxdg − lim ftsxd f sxdg − lim tsxd ? lim f sxd
x l a x l a x l a x l a
− lim
x l a
c ? lim
xl a
f sxd
− c lim
xla
f sxd (by Law 7) ■
Proof of Law 2 Using Law 1 and Law 3 with c − 21, we have
lim f f sxd 2 tsxdg − lim f f sxd 1 s21dtsxdg − lim f sxd 1 lim s21dtsxd
x l a x l a xl a x l a
− lim
x l a
f sxd 1 s21d lim
x l a
tsxd − lim
x l a
f sxd 2 lim
x l a
tsxd
■
Proof of Law 5 First let us show that
lim
x l a
1
tsxd − 1 M
To do this we must show that, given « . 0, there exists . 0 such that
if 0 , | x 2 a | , then Z
1
tsxd 2 1 M
Z , «
Observe that Z
1
tsxd 2 1 M
Z − | M 2 tsxd |
| Mtsxd |
We know that we can make the numerator small. But we also need to know that the
denominator is not small when x is near a. Since lim x l a tsxd − M, there is a number
1 . 0 such that, whenever 0 , | x 2 a | , 1, we have
| tsxd 2 M | | , M |
2
and therefore | M | − | M 2 tsxd 1 tsxd | < | M 2 tsxd | 1 | tsxd |
| , M |
1
2
| tsxd |
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