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Unformatted text preview: q log x x log 1
q log x x log
x
x
1 q log x x log x q log x x log x
Now given any sufficiently large number x, we can apply the mean value theorem to the function log on
the interval x q log x, log x to obtain a number f x between x q log x and log x at which the derivative
of the function log is equal to
log x log x q log x
x x q log x
and we can express f x in the form x g x where 0 g x
q log x. Thus
log x log x q log x
1
q log x
x gx
and we conclude that
q log x
q log x x log x q log x x log x q log x x
x gx
q log x 1 x
gx x Therefore
q log x
x q log x x log 1 q log x
q log x x x
gx x x
q log x 1
1 log x 2 q2 x
q 2 log x 2
log x
x q log x
1 qx
Ý. This completes our consideration of the case q 0.
and the latter expression approaches 0 as x Finally we consider the case q 0. We write p q. Then p 0 and we want to show that
p log x x
lim 1 1
1.
p
x
xÝx
We need to show that
p log x
x log 1
p log x 0
x
as x Ý. Now whenever x is sufficiently large we have
p log x
x p log x
x log 1
p log x x log
x
x and for some number g x satisfying 0 gx x log x p log x
p log x we have 254 p log x
x log x p log x x log 1 p log x
x px log x
xg x p log x p log x and so
p log x
x p log x p log x 1 x
xg x p log x x log 1 1 x
x p log x log x 2
x
log x
p x p
1 Ý. and the latter expression approaches 0 as x
6. Suppose that we want to evaluate the limit 3
.
1 x x2
Would it be correct to use L’Hôpital’s rule by taking f x 3 and g x 1 x x 2 for every x and then
arguing that
fx
fx
3
lim
lim
lim
lim 0
0?
x Ý 1 x x2
xÝgx
xÝg x
x Ý 1 2x
The answer is certainly correct, but is the reasoning correct? Have we made a valid use of L’Hôpital’s rule?
lim
xÝ Solution: Yes, we have made a valid application of L’Hôpital’s rule. The fact that the denominator of
the fraction
3
1 x x2
approaches Ý as x Ý is all we need; even though most elementary calculus books give the impression
that L’Hôpital’s rule requires both the numerator and the denominator to approach Ý.
7. Evaluate the limits
sin x
lim e x
xÝ and sin x
lim xe
.
x Ý log x Should L’Hôpital’s rule be used to evaluate these limits?
L’Hôpital’s rule would say that
sin x
e sin x cos x
lim e x lim
xÝ
xÝ
1
provided that the latter limit exists. But it doesn’t. So L’Hôpital’s rule can’t be used here. Note,
however that
sin x
lim e x 0.
xÝ
L’Hôpital’s rule would say that
sin x
e sin x x e sin x cos x
lim xe
lim
1
x Ý log x
xÝ
x and the latter limit does not exist. Nor does the original limit. In fact, the graph
sin x
y xe
log x
is 255 30
25
20
15
10
5
10 20 x 30 40 50 10 The Exponential and Logarithmic Functions
Some Exercises on the Exponential and Logarithmic
Functions
1. Given that a is a positive number and that f x a x for every real number x, prove that
f x a x log a
for every number x. In other words, prove that the number we called a in the preceding sections is ju...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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