limit of ln as x goes to infinity


by Cacophony
Tags: infinity, limit
Cacophony
Cacophony is offline
#1
Mar7-12, 10:25 PM
P: 41
1. The problem statement, all variables and given/known data
lim (lnx)^2/x
x-->infinity


2. Relevant equations

none

3. The attempt at a solution

=5lnx/x * (1/lnx)/(1/lnx)

=5/(x/lnx)

How do I calculate x/lnx?
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ehild
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#2
Mar7-12, 11:37 PM
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Quote Quote by Cacophony View Post

3. The attempt at a solution

=5lnx/x * (1/lnx)/(1/lnx)

=5/(x/lnx)

How do I calculate x/lnx?
I do not follow you. What have you done?

ehild
Mark44
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#3
Mar7-12, 11:40 PM
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Quote Quote by Cacophony View Post
1. The problem statement, all variables and given/known data
lim (lnx)^2/x
x-->infinity


2. Relevant equations

none

3. The attempt at a solution

=5lnx/x * (1/lnx)/(1/lnx)
Where did the 5 come from? In fact, where did any of this come from? What you have makes zero sense to me.

Also, since you haven't taken the limit yet, you should not get rid of the "lim" symbol.
Quote Quote by Cacophony View Post

=5/(x/lnx)

How do I calculate x/lnx?
This is a problem that is suited to L'Hopital's Rule. Have you covered it yet?

checkitagain
checkitagain is offline
#4
Mar8-12, 12:12 PM
P: 99

limit of ln as x goes to infinity


Quote Quote by Cacophony View Post
1. The problem statement, all variables and given/known data
lim (lnx)^2/x
x-->infinity
Cacophony,

what you typed is equivalent to:

[tex]\displaystyle \lim_{x\to \infty}\dfrac{[ln(x)]^2}{x}[/tex]



Did you intend

[tex]\displaystyle \lim_{x\to \infty}[ln(x)]^{\frac{2}{x}} \ ?[/tex]


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