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Unformatted text preview: case 2
we approached the problem using L’Hôpital’s rule and again using the mean value theorem. This 251 time we shall use only the mean value theorem. For each x 0 we choose a number f x between
x and x 1 such that
log f x 1
log x 1 log x
.
fx
If we define
log u 1
gu
u
for u 0 then since
1 log u 2 log u 1
1 log u
log u 2
0
gu
u2
u2
whenever u exp 1 we have
log x 1
log x 1 1
log x 1 log x
x
x1
whenever x exp 1 and so
lim log x 1
xÝ log x 0. Now we discuss the limit
lim
xÝ log x 1
log x x lim exp x log
xÝ log x 1
log x lim exp x log log x 1
xÝ log log x We look first at the limit
lim x log log x 1
xÝ lim
xÝ log log x log log x 1 log log x 1
x Once again, L’Hôpital’s rule can be used but the mean value theorem presents an attractive
alternative. For each x 0 we choose a number f x between x and x 1 such that
1
log log x 1 log log x
f x log f x
from which we see (in the same way we argued in the preceding exercises) that
log log x 1 log log x
x
x
1
x log x
x 1 log x 1
x
and so
log log x 1 lim
xÝ log log x 1
x 0 and we conclude that
lim
xÝ
Now we consider the limit
log x 1
lim
xÝ
log x x log x log x 1
log x x exp 0 1. lim exp
xÝ x log x log lim exp
xÝ x log x log x 1
log x log log x 1 log log x We look first at the limit
lim
xÝ x log x log log x 1 log log x 252 lim
xÝ log log x 1
1
x log x log log x Arguing as above we see that
x log x
x log x log log x 1 log log x x log x
x 1 log x 1 1
x log x and so
log log x 1 lim
xÝ log log x 1 1
x log x and we conclude that
log x 1
log x lim
xÝ
4. a. Prove that if 0 a b and x x log x exp 1 e. 1 then bx ax
xb x 1 .
ba
Suppose that x 1. For every positive number t we define
f t tx
and we apply the mean value theorem to f on the interval a, b .
xa x 1 b. Prove that
lim log x 1
xÝ log x log x Ý. log x Solution: From part a we know that, whenever x is sufficiently large,
log x log x log x 1 log x
log x 1 log x 1 log x log x
log x log x log x 1 log x 1 Thus
log x 1
log x 1 log x log x log x
log x log x 1
x log x log x 1 log x log x log x log x log x log x . To see that the latter expression clearly approaches Ý as x Ý, we write it as
log x 1 log x
exp log x log log x
.
x
1
x It is easy to see that
lim
xÝ log x 1 log x 1 1
x and since log log x 2 for all sufficiently large x we have
exp log x log log x
exp 2 log x
x
x x for all x sufficiently large and so
lim
xÝ exp log x log log x
x Ý. c. Prove that
lim log x 1
xÝ log log x log x log log x 0. 5. Prove that if q is any given number then
lim x q 1
xÝ q log x
x x 1 Solution: The assertion we want to prove is obviously true when q 0. We now consider the case
253 q 0.
We need to show that
q log x
x q log x x log 1
as x 0 Ý. Since
lim
xÝ log x
x 0 we know that
q log x
0
x
whenever x is sufficiently large. Thus, for x sufficiently large we have
q log x
x q log x...
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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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