Question

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

Answer 1

Bad way to start. Get a common denominator: [(1*lnx)-(1*(x-1))]/[(x-1)lnx]

Answer 2

Let a=1/0

a-a=0. Therefore, your answer is 0.

Answer 3

I got -1/2, anyone agree?

Answer 4

you got it right. It is -1/2. Using hopitals rule.

Answer 5

@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.

Answer 6

Actually, this is probably the best way-> http://www.wolframalpha.com/input/?i=lim+x-%3E1+%28%28 [1%2F%28x-1%29]-[1%2Flnx]%29%29

Answer 7

thanks, yeah using hopitals rule twice

Answer 8

oops, the url broke

Answer 9

that wolframalpha thing is realllly nice

Answer 10

Oh damn, you're right. Man I suck at math.

Answer 11

no, indeterminants are just really weird

Answer 12

Harwin -- easy mistake to make. My students do it all the time.

Answer 13

I was joking.

Answer 14

wolframalpha.com (in case you haven't heard of it already)--it helps a lot with understanding IMO

Answer 15

k thanks Esperantist, i'll check it out

Answer 16

If anyone's still around I have sin(x)/[1-cos(x)], I got -1/2 again, is that right?

Answer 17

what does it approach to?