Find

Can I do this:

where l'Hopital's rule is applied to the numerator only. Then reapply l'Hopital's rule again:

Lemme know if my divide and conquer approach with l'Hopital's rule is legit.

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- February 22nd 2011, 02:37 PMcrossbonel'Hopital's rule
Find

Can I do this:

where l'Hopital's rule is applied to the numerator only. Then reapply l'Hopital's rule again:

Lemme know if my divide and conquer approach with l'Hopital's rule is legit. - February 22nd 2011, 04:20 PMskeeter

the limit DNE - February 22nd 2011, 04:32 PMcrossbone
so divide and conquer with l'Hopital and reapplying it is not legal?

- February 22nd 2011, 06:18 PMskeeter
I would say not ... the original limit does not exist.

limit as x to 0 (x*log(x))/(log(1& #43;2x)) - Wolfram|Alpha - February 22nd 2011, 07:35 PMcrossbone
Here's another problem:

- February 22nd 2011, 07:43 PMProve It
Rewrite it as , which is now of the form so you can now use L'Hospital's Rule...

- February 22nd 2011, 07:57 PMcrossbone
I knew that but this is what I got:

keep differentiating until a-n=b<0:

where

Is this right? - February 22nd 2011, 08:49 PMProve It
The derivative of is not .

You need to use the chain rule... - February 22nd 2011, 09:27 PMAllanCuz
What exactly does this approach violate? I'm looking at the rules for L`Hopitals and I don't quite see where this goes wrong, but it obviously does violate something: http://bradley.bradley.edu/~delgado/121/Lhopital.pdf

Any ideas? - February 22nd 2011, 10:01 PMFernandoRevilla

Right. With this, you prove

Quote:

Then reapply l'Hopital's rule again:

Lemme know if my divide and conquer approach with l'Hopital's rule is legit.

But now, you find:

Fernando Revilla - February 22nd 2011, 11:11 PMcrossbone