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anonymous
 5 years ago
hey...lim as x>1 of ([1/(x1)][1/lnx])...I know it is [(1/0)(1/0)], what do I do from there?
anonymous
 5 years ago
hey...lim as x>1 of ([1/(x1)][1/lnx])...I know it is [(1/0)(1/0)], what do I do from there?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Bad way to start. Get a common denominator: [(1*lnx)(1*(x1))]/[(x1)lnx]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let a=1/0 aa=0. Therefore, your answer is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 1/2, anyone agree?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you got it right. It is 1/2. Using hopitals rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@Harwin: No can do. 1/0 = infinity. Infinity  infinity is one of the 7 indeterminate forms. it's not 0. Esperantist is correct. Get a common denominator. Then it may be possible to use l'Hopital's rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, this is probably the best way> http://www.wolframalpha.com/input/?i=lim+x%3E1+%28%28 [1%2F%28x1%29][1%2Flnx]%29%29

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks, yeah using hopitals rule twice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that wolframalpha thing is realllly nice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh damn, you're right. Man I suck at math.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, indeterminants are just really weird

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Harwin  easy mistake to make. My students do it all the time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wolframalpha.com (in case you haven't heard of it already)it helps a lot with understanding IMO

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k thanks Esperantist, i'll check it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If anyone's still around I have sin(x)/[1cos(x)], I got 1/2 again, is that right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what does it approach to?
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