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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.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.