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SECTION 8.3
R E L AT I V E R AT E S O F G R O W T H
SECTION 8.3
RELATIVE RATES OF GROWTH
Learning Target:
–I can compare rates of growth (grows faster/slower/at
the same rate)
COMPARING RATES OF GROWTH
• Definition: Faster, Slower, Same-rate Growth as x  
• Let f(x) and g(x) be positive for x sufficiently
large.
1. f grows faster than g (and g grows slower than f)
f ( x)
 , or equivalently, if lim g ( x)  0.
as x   if lim
x 
g ( x)
x 
f ( x)
2. f and g grow at the same rate as x   if
f ( x)
lim
 L  0.
x  g ( x )
EXAMPLE 1: (GROWS FASTER)
Shows that the function ex grows faster than x2 as x
 .
e

lim 2 
x  x

x
L' Hop
e

lim

x  2 x

L' Hop
x
e

lim



x  2
2
x
Therefore, ex grows
faster than x2.
EXPLORATION
1. Show that ax, a>1, grows faster than x2 as x  
L' Hop x
a
a ln a 

lim 2 
lim

x  x

 x  2 x
x
L' Hop
x
2
a (ln a )
lim


x 
2
x grows
Therefore,
a
2. Show that 3x grows faster than 2x as x  
faster than x2.
x
x
3
Therefore, 3x grows
3
lim x  lim   
x.
faster
than
2
x 
x 
2
2
3. If a > b > 1, show that ax grows faster than bx as x  
x
a
lim x
x  b
x
Therefore, ax grows
a
 lim   
x.
faster
than
b
x 
Because a/b > 1
b
EXAMPLE 2: (GROWS SLOWER)
Show that ln x grows slower than:
a)

ln x 
lim

x  x

x as x
L' Hop
1
1
x
 lim
lim
0
x  1
x  x
Therefore, ln x grows
slower than x.
b)

ln x 
lim 2 
x  x

x2 as x
L' Hop
1
1
x
lim
 lim 2  0
x  2 x
x  2 x
Therefore, ln x grows
slower than x2.
EXAMPLE 3:
(GROWS AT THE SAME RATE)
Let a and b be numbers greater than 1. Show that logax and
logbx grow at the same rate as x   .
ln b
ln x ln a
log a x
 lim
 lim
lim
0
x  ln a
x  ln x ln b
x  log x
b
Therefore, they grow at
the same rate.
DEFINITION
• Transitivity of Growing Rates – If f grows at the
same rate as g as x   and g grows at the same rate
as h as x   , then f grows at the same rate as h as
x
**Note: Growing at the same rate is a transitive
relation.
EXAMPLE 4:
Put the following functions in order from SLOWEST
x
2
x
x
to FASTEST: 2 , x , (ln 2) , e
x
x
x
2
2
2 ln 2 
2 (ln 2)

lim 2 
lim
lim



x  x
x 
x 
2
x
2


x grows
Therefore,
2
x
x
2.
faster
than
x
2
 2 
x grows
Therefore,
2

lim
lim



x 
x  (ln 2) x
faster than (ln2)x.
 ln 2 
x
e
 e 
lim
 lim



x
x  (ln 2)
x 
 ln 2 
x
Therefore, ex grows
faster than (ln2)x.
EXAMPLE 4, CONTINUED
x

lim

x
x  (ln 2)

2
x
e
lim x
x  2
2x
lim
x  (ln 2) x ln(ln 2)
Therefore, (ln2)x grows
2.
faster
than
x
2
lim
0
x  (ln 2) x ln(ln 2) 2
x
e
 lim   
x 
2
2
Therefore, ex grows
faster than 2x.
x
x
x , (ln 2) ,2 , e
x
8.3 HOMEWORK
#3, 6, 9, 12, 17, 24, 25, 27, 30, 46 - 51
QUIZ DAY 
After you complete your quiz, please complete pages
12 – 14… Integral Refreshers. I will be checking off
8.3 and the Integral Refreshers WS tomorrow in
class.