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limit of ln as x goes to infinity

 
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Mar7-12, 10:25 PM   #1
 

limit of ln as x goes to infinity


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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Mar7-12, 11:37 PM   #2

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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
Mar7-12, 11:40 PM   #3
 
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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 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?
Mar8-12, 12:12 PM   #4
 

limit of ln as x goes to infinity


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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