Unformatted text preview: uÝ
lim exp u 100
exp u 2 follows from the composition theorem for limits.
We now discuss the limit
lim sin x x /2 tan x lim exp tan x log sin x
x /2 We look first at the limit 249 lim tan x log sin x lim x /2 x /2 log sin x
cos x sin x lim sin x log sin x
cos x lim x /2 x /2 log sin x
cos x lim x /2 Applying L’Hôpital’s rule to the latter limit we obtain
log sin x
lim
lim
cos x
x /2
x /2 cot x
0
sin x
and so it follows from the composition theorem for limits that
lim sin x tan x exp 0 1
x /2 Now we discuss the limit
lim x
xÝ log x
x . Since this limit can be expressed as
log x
x lim exp
xÝ 2 and since
log x
x lim
xÝ 2 0 we have
lim x
xÝ log x
x exp 0 1. Now we discuss the limit
lim
xÝ x log x
x 1 log x1 lim
xÝ log x exp 2 log x 1 exp lim exp
xÝ log x 2 log x 1 2 2 We look first at
lim log x 1
xÝ 2 log x 2 It is possible to find this limit using L’Hôpital’s rule by expressing it as
lim log x 1
xÝ log x log x 1 log x lim
xÝ log 1 1
x 1
log x 2 x which gives us lim
xÝ x
x 1 1
x2 1
log x 2 x 2 2 x 1
x 2 x which can be seen easily to be 0. Almost every student will want to do the problem this way. On the
other hand, one can also use the mean value theorem to choose a number that we shall call f x
between x and x 1 such that
log x 1 2 log x 2
2 log f x
log x 1 2 log x 2
.
x1 x
fx
Since the function g defined by the equation
log u
gu u
for u e, having a negative derivative, must be decreasing we have 250 1 2 log f x
fx log x
x log x 1
x1 1 and it follows that
lim
xÝ 2 log f x
fx 0. From either method we deduce that
lim
xÝ x log x
x 1 log x1 2. Given that 0, evaluate the limit
lim x
xÝ exp 0 1. 2x 2 2x 1
2x 2 . Solution: Although L’Hôpital’s rule can be used to solve this problem, it is not the best approach.
Instead we should define
f x x
for all x 0 and, for a given positive number x, apply the mean value theorem to the function f on the
interval 2x 1, 2x 2 to choose a number c between 2x 1 and 2x 2 such that
f 2x 2 f 2x 1
f c c 1
2x 2 2x 1
2x 2
2x 1
In the event that 1 we know from the fact that 2x 1 c 2x 2 that
2x 1 1 c 1 2x 2 1
and so
x 2x 1 1
2x 2 2x 1
x 2x 2 1
x
2x 2
2x 2
2x 2
which we can express as
2x 1
x
x
2x 1 1 x 2x 2
2x 2
2x 2 2x 2
2x 2
and we deduce from the sandwich theorem that
2x 2 2x 1
lim x
xÝ
2
2x 2
In the event that 0 1 the inequalities are reversed but the result is the same. Try to handle this case
yourself.
3. Evaluate the limits
lim log x 1
xÝ
lim
xÝ log x 1
log x log x log x 1
log x x lim
xÝ x log x For the latter two limits, check the limit value with Scientific Notebook.
The limit
lim log x 1
xÝ lim log x 1
xÝ 2 log x is an extension of the limit
log x 2 that is discussed above. We understand to be a given constant in this exercise. In the...
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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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