What is the solution for a limit problem with an indeterminate form of 0/0?

[Macleef]

 

[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.

 

[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.

 

[Dick]

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.

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.