amaths

1. limx->2 [√(x+2)-√(3x-2)]/[√(5x-1)-√(4x+1)]
=limx->2 [√(x+2)-√(3x-2)][√(5x-1)+√(4x+1)]/[(5x-1)-(4x+1)]
=limx->2 [√(x+2)-√(3x-2)][√(5x-1)+√(4x+1)]/(x-2)
=limx->2 [(x+2)-(3x-2)][√(5x-1)+√(4x+1)]/(x-2)[√(x+2)+√(3x-2)]
=limx->2 [-2(x-2)][√(5x-1)+√(4x+1)]/(x-2)[√(x+2)+√(3x-2)]
=-2limx->2 [√(5x-1)+√(4x+1)]/[√(x+2)+√(3x-2)]
=-2(√9+√9)/(√4+√4)
=-2(6/4)
=-3

2. limh->0 [(1+h)^(1/3)-1]/h
=limh->0 [(1+h)^(1/3)-1][(1+h)^(2/3)+(1+h)^(1/3)+1]/(h)[(1+h)^(2/3)+(1+h)^(1/3)+1]
=limh->0 1/[(1+h)^(2/3)+(1+h)^(1/3)+1]
=1/(1+1+1)
=1/3

3. limx->1 [x^(1/3)-1]/(√x-1)
=limx->1 [x^(1/3)-1](√x+1)/(x-1)
=limx->1 (x-1)(√x+1)/(x-1)[x^(2/3)+x^(1/3)+1)
=limx->1 (√x+1)/[x^(2/3)+x^(1/3)+1)
=(1+1)/(1+1+1)
=2/3

P.S.
For Q1, you should rationalize it twice to get the answer because you find rationalize it once do not work.
For Q2, you should use both the identities (x-y)(x+y)≡x^2-y^2 and (x±y)(x^2∓x+1)=x^3±1
For Q3, you should use the skills from Q1 and Q2 altogether the get the answer.

我也做了,不過慢很多,所以作為意見!
http://www.photo-host.org/img/704528screenhunter_01_jun._30_23.08.gif
http://www.photo-host.org/img/999378screenhunter_02_jun._30_23.09.gif
http://www.photo-host.org/img/421147screenhunter_03_jun._30_23.09.gif