Solving Limit: x.log(x) - Indetermination

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Homework Help Overview

The problem involves evaluating the limit of the expression x.log(x) as x approaches positive infinity. The original poster notes that while log(x) tends to infinity, the product x.log(x) presents an indeterminate form, which they seek to resolve.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants discuss the nature of the limit, questioning whether the form x.log(x) truly represents an indeterminate case. Others suggest rewriting the expression to apply l'Hopital's rule, while some counter that this approach is not valid due to the resulting form.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the limit and the validity of applying l'Hopital's rule. There is no explicit consensus on the correct approach to take.

Contextual Notes

Participants are navigating the definitions of indeterminate forms and the conditions under which l'Hopital's rule can be applied, indicating some uncertainty about the problem's setup.

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Homework Statement



The problem is

Limit [x.log(x)]
x->+oo

Homework Equations



Consider Log being a logarithm of base 10

This will tend to +oo, but x.log(x) will become (oo).(oo) which is an indetermination I need to know how to solve the indetermination in this case

The Attempt at a Solution



Thanks for all help
 
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[itex]\infty * \infty[/itex] is NOT an indeterminate form. If you have two quantities that are getting larger and larger, their product will, too. The limit is infinity.
 
Bear in mind that
[tex]x\log(x) = \frac{\log(x)}{\frac{1}{x}}.[/tex]
Given that form, you can apply l'Hopital's rule.
 
No, you can't. That becomes "infinity over 0" which, again, is NOT an "indeterminant". L'Hopilal's rule does not apply and you don't need it.
 
HallsofIvy said:
No, you can't. That becomes "infinity over 0" which, again, is NOT an "indeterminant". L'Hopilal's rule does not apply and you don't need it.

My apologies; read it too quickly and thought we were talking about a limit as x->0+.
 

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