anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Bad way to start. Get a common denominator: [(1*lnx)-(1*(x-1))]/[(x-1)lnx]
anonymous
  • anonymous
Let a=1/0 a-a=0. Therefore, your answer is 0.
anonymous
  • anonymous
I got -1/2, anyone agree?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
you got it right. It is -1/2. Using hopitals rule.
anonymous
  • anonymous
@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
  • anonymous
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
anonymous
  • anonymous
thanks, yeah using hopitals rule twice
anonymous
  • anonymous
oops, the url broke
anonymous
  • anonymous
that wolframalpha thing is realllly nice
anonymous
  • anonymous
Oh damn, you're right. Man I suck at math.
anonymous
  • anonymous
no, indeterminants are just really weird
anonymous
  • anonymous
Harwin -- easy mistake to make. My students do it all the time.
anonymous
  • anonymous
I was joking.
anonymous
  • anonymous
wolframalpha.com (in case you haven't heard of it already)--it helps a lot with understanding IMO
anonymous
  • anonymous
k thanks Esperantist, i'll check it out
anonymous
  • anonymous
If anyone's still around I have sin(x)/[1-cos(x)], I got -1/2 again, is that right?
anonymous
  • anonymous
what does it approach to?

Looking for something else?

Not the answer you are looking for? Search for more explanations.