# Just another limit problem

1. Apr 8, 2008

2. Apr 8, 2008

### Mathdope

You should have 0/0 not 6/0. In any case, try a trick similar to what you did with the radical in the denominator.

3. Apr 8, 2008

### workerant

Well if you plug the four in initially you get:

2-2/(3-3)=0/0

Your trick did not work nor does using the other radical as Mathdope suggests.

But, never fear, 0/0 limits call for one man.....

L'Hopital!

Since it is indeterminate form, it is eligible for L'Hopital's rule. That should make it work.

4. Apr 8, 2008

### Dick

Multiplying by 2+sqrt(x) does so work. You can then cancel the factor that's going to zero in the numerator and the denominator.