Thomas Calculus 13th [Solutions]

Section 7.4 Relative Rates of Growth 531
ln x
log3
x
ln 3 1 1
5. (a) same, lim lim lim
x
ln x
x
ln x
x
ln 3 ln 3
2
2
(b) same, lim
ln 2x
x
lim
ln
1
x
x
x
x
1
1
ln x
2
(c) same,
x
lim lim lim
1 1
x
ln x
x
ln x
x
2 2
1 1/2
1/2
x
x
2
(d) faster, lim lim
x
lim lim
x
lim
1
x
x
x
x
x 2 x
x
x x
(e) faster, lim
x
lim
1
lim x
x
ln ln 2
x
ln x
x x
1
x
(f ) same, lim 5ln x
lim 5 5
(g) slower,
(h) faster,
x
x
ln x
1
x
x
1
ln x
x
x ln x
lim lim 0
x
e e
x
ln x
x x
lim lim lim xe
x
1
x
x
6. (a) same,
(b) same,
ln x
2
2 ln 2
2
2
log x
lim lim
1
lim
ln x 1
lim
2ln x 1
lim 2
2
x
ln x
x
ln x ln 2
x
ln x ln 2
x
ln x ln 2
x
ln 2
ln10x
10
log1010x
ln10 ln10x
10x
x x x
1
x
lim lim
1
lim
1
lim
1
lim 1
1
x
ln
x
ln ln10
x
ln ln10
x
ln10
x
ln10
(c) slower,
(d) slower,
x
x
1
x
1
ln x
x x (ln x)
lim lim 0
1
x
2
1
ln x 2
x x
lim lim 0
x
ln
x 2ln x x x
1
x
ln x
x
ln x
x
ln x
x x
(e) faster, lim lim 2 lim 2 lim 2 lim x 2
1
e
ln x
x
x
e ln x
(f ) slower, lim lim 1
0
(g) slower,
(h) same,
x
x
1/ x
ln x
1
x
ln(ln x) 1
x
ln x
x x
ln x
lim lim lim 0
2
2x
5
1
x
ln(2x
5) 2x
2
x
ln x
x x
2x
5
x
2
x
lim lim lim lim lim 1 1
x
x
7. lim
e
x/2
x
e
ln x
x
x
lim
x/2
e
x
e grows faster then
e x/2 ; since for
e
x e we have ln x e and
lim lim (ln x ) grows faster then e ; since x ln x for all x 0 and
x
e
x
x
lim
x
lim
x
x (ln x)
x
x/2 ,
x ,(ln )
x ,
x
x
ln x
x
e
x
x
ln x
e e x x so the order is d, a, c, b
x
x
x
x
x grows faster then (ln x ) . Therefore, slowest to fastest are:
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