334
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
We have already shown that, for a > 1 for example, an tends to +00 (resp. 0)
as n E IE tends to +00 (resp. -oo)j the function being increasing, (1) and (2)
follow trivially.
At the end points of their interval of definition x > 0 the power functions
tend to easily calculated limits. First,
(5.3)
lim X S = { 0
x-+O,x>O
+00
if s > 0
if s < O.
Assume for example that s > O. We have to show that, for all r > 0, we
have X S < r for x > 0 sufficiently smallj but on writing r in the form as, this
inequality is equivalent to x < a since the function X S is strictly increasing.
The case s < 0 reduces to the previous one, since X S = 1/x- s .
We have the same results, inverted, when x increases indefinitely:
(5.4)
1 ·
S
{+OO
1m x =
X-++OO
0
if s > 0
if s < O.
If s > 0 we have to verify that X S > M for x large, which is clear since this
relation is equivalent to x > MIls.
The log functions also have a simple behaviour:
(5.5)
lim loga x = +00,
X-++oo
lim loga x = -00 if a > 1.
X-+O
It is enough to establish the first, the second reducing to this because
log x = - loge 1 I x). Now we know that the function loga x is increasing, > 0
for x > 1, and that loga (xn) = n loga x. This relation shows that loga x takes
arbitrarily large values for x > 1, and since it increases, it is forced to grow
indefinitely. For 0 < a < lone inverts the results, because loglla x = -loga x.
Now let us try to compare the growth of these various functions as x increases indefinitely. The results are very simple. First, the following formulae
for those that grow indefinitely with x:
(5.6)
aX
o(b X ) when x ...... +00 if 1 < a < bj
(5.7)
X S
o(a X )
if s > 0,1 < aj
(5.8)
X S
o(xt)
if 0 < s < tj
(5.9)
loga x
o(X S )
if s > O,a > O.
The first means that the ratio aX Ib x
ex, where c = alb, tends to 0 at
infinityj obvious since c < 1. The third means that X S Ixt = x s - t tends to OJ
obvious since s - t < 0.
To establish (7), one may assume that s is an integer p (if not, choose a
p> s, whence X S < x P ). Putting a = b P , where again b > 1, we have only to
show that x P IbP x = (xlbX)P tends to 0. So it is enough to examine the ratio
IV - Powers, Exponentials, Logarithms, Trigonometric Functions
We have already shown that, for a > 1 for example, an tends to +00 (resp. 0)
as n E IE tends to +00 (resp. -oo)j the function being increasing, (1) and (2)
follow trivially.
At the end points of their interval of definition x > 0 the power functions
tend to easily calculated limits. First,
(5.3)
lim X S = { 0
x-+O,x>O
+00
if s > 0
if s < O.
Assume for example that s > O. We have to show that, for all r > 0, we
have X S < r for x > 0 sufficiently smallj but on writing r in the form as, this
inequality is equivalent to x < a since the function X S is strictly increasing.
The case s < 0 reduces to the previous one, since X S = 1/x- s .
We have the same results, inverted, when x increases indefinitely:
(5.4)
1 ·
S
{+OO
1m x =
X-++OO
0
if s > 0
if s < O.
If s > 0 we have to verify that X S > M for x large, which is clear since this
relation is equivalent to x > MIls.
The log functions also have a simple behaviour:
(5.5)
lim loga x = +00,
X-++oo
lim loga x = -00 if a > 1.
X-+O
It is enough to establish the first, the second reducing to this because
log x = - loge 1 I x). Now we know that the function loga x is increasing, > 0
for x > 1, and that loga (xn) = n loga x. This relation shows that loga x takes
arbitrarily large values for x > 1, and since it increases, it is forced to grow
indefinitely. For 0 < a < lone inverts the results, because loglla x = -loga x.
Now let us try to compare the growth of these various functions as x increases indefinitely. The results are very simple. First, the following formulae
for those that grow indefinitely with x:
(5.6)
aX
o(b X ) when x ...... +00 if 1 < a < bj
(5.7)
X S
o(a X )
if s > 0,1 < aj
(5.8)
X S
o(xt)
if 0 < s < tj
(5.9)
loga x
o(X S )
if s > O,a > O.
The first means that the ratio aX Ib x
ex, where c = alb, tends to 0 at
infinityj obvious since c < 1. The third means that X S Ixt = x s - t tends to OJ
obvious since s - t < 0.
To establish (7), one may assume that s is an integer p (if not, choose a
p> s, whence X S < x P ). Putting a = b P , where again b > 1, we have only to
show that x P IbP x = (xlbX)P tends to 0. So it is enough to examine the ratio