Using L'Hospital's rule with roots and log functions

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression involving the square root and natural logarithm as x approaches 0, specifically lim x -> 0 [ \sqrt{x}*ln(x) ]. The subject area includes calculus concepts, particularly limits and L'Hospital's rule.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of L'Hospital's rule, with some suggesting rewriting the expression as a quotient to fit the criteria for the rule. Questions arise regarding the manipulation of exponents and the validity of rewriting terms.

Discussion Status

The discussion is active, with participants providing different perspectives on how to approach the limit. Some guidance has been offered regarding rewriting the expression, but there is no explicit consensus on the method to apply.

Contextual Notes

There is confusion regarding the application of L'Hospital's rule and the manipulation of the expression involving exponents. Participants are questioning the assumptions made in rewriting the terms.

fiziksfun
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Can someone help me use L'HOP to determine

lim x -> 0 [ [tex]\sqrt{x}[/tex]*ln(x) ]

? I'm confused!
 
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fiziksfun said:
Can someone help me use L'HOP to determine

lim x -> 0 [ [tex]\sqrt{x}[/tex]*ln(x) ]

? I'm confused!
L'Hopital's rule doesn't apply here. One can only apply L'Hopital's rule for a limit of a quotient and only then when the limit is undefined.
 


Write it as ln(x)/x^(-1/2). Now it's a quotient and infinity/infinity. Looks good for l'hop.
 


Dick said:
Write it as ln(x)/x^(-1/2). Now it's a quotient and infinity/infinity. Looks good for l'hop.
*Hangs head in shame and shuffles back into the Physics section*
 


Dick said:
Write it as ln(x)/x^(-1/2). Now it's a quotient and infinity/infinity. Looks good for l'hop.

why can i rewrite x^(1/2) as x^(-1/2) ? I don't understand.
 


fiziksfun said:
why can i rewrite x^(1/2) as x^(-1/2) ? I don't understand.

You can't, but you can write [tex]x^{1/2}[/tex] as [tex]\frac{1}{x^{-1/2}}[/tex], which is what Dick has done above.
 


fiziksfun said:
why can i rewrite x^(1/2) as x^(-1/2) ? I don't understand.
You can't rewrite x^(1/2) as x^(-1/2), but you can rewrite it as,

[tex]x^{1/2} = \frac{1}{x^{-1/2}}[/tex]

as Dick suggests.

Edit: Get out of my head cristo :-p
 


Hootenanny said:
Edit: Get out of my head cristo :-p

:wink:
 

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