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fiziksfun
Oct7-08, 09:00 AM
Can someone help me use L'HOP to determine

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

??? I'm confused!!

Hootenanny
Oct7-08, 09:05 AM
Can someone help me use L'HOP to determine

lim x -> 0 [ \sqrt{x}*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.

Dick
Oct7-08, 09:32 AM
Write it as ln(x)/x^(-1/2). Now it's a quotient and infinity/infinity. Looks good for l'hop.

Hootenanny
Oct7-08, 09:40 AM
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*

fiziksfun
Oct7-08, 12:16 PM
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.

cristo
Oct7-08, 12:17 PM
why can i rewrite x^(1/2) as x^(-1/2) ??? I don't understand.

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

Hootenanny
Oct7-08, 12:19 PM
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,

x^{1/2} = \frac{1}{x^{-1/2}}

as Dick suggests.

Edit: Get out of my head cristo :tongue2:

cristo
Oct7-08, 12:22 PM
Edit: Get out of my head cristo :tongue2:

:wink: