1) lim [ ( x^2+2x-15)/(x^2-9)] , x->3
= (9+6-15)/(9-9)
= 0/0 (indeterminate form)
Applying L'Hopital's rule:
= lim [ (2x+2)/(2x)] , x->3
= (2(3)+2)/(2(3))
= 8/6
= 4/3

2) lim { [ 2z]/ [ 4 + z )^1/2 - ( 4-z)^1/2]} , z->0
= [2(0)]/[4 + 0 - 4]
= 0/0 (indeterminate form)
Applying L'Hopital's rule:
= lim { 2/[1/2(4 + z)^1/2 + 1/2(4-z)^1/2]} , z -> 0
= 2/[1/2(4)^(1/2) + 1/2(4)^(1/2)]
= 2/[1/4 + 1/4]
= 2/(1/2)
= 4

3) lim{[2-x^1/2]/[3-(2x+1)1/2]},x->4
Substitute x = 4:
= [2-4^(1/2)] / [3 - 9^(1/2)]
= [2-2] / [3 - 3]
= 0/0 (indeterminate form)
Applying L'Hopital's rule:
= lim{[-1/(2sqrt(x))]/[-1/(2sqrt(2x+1))]}, x->4
= [-(1/2sqrt(4))]/[-(1/2sqrt(2(4)+1))]
= [-1/4]/[-1/2sqrt(9)]
= -1/4 / -1/(23)
= -1/4 / -1/6
= -1/4 -6
= 3/2

4) lim{[x+27]/[(x)^(1/3)+3]}, x-> -27
Substitute x = -27:
= [-27 + 27] / [(-27)^(1/3) + 3]
= 0 / [(-27)^(1/3) + 3]
= 0 / [-3 + 3]
= 0/0 (indeterminate form)
Applying L'Hopital's rule:
= lim {[1/(3x^(-2/3))]/[1/(3x^(-2/3))]}, x -> -27
= 1/(3(-27)^(-2/3))/1/(3(-27)^(-2/3))
= 1/(3(1/9))/1/(3(1/9))
= 1/3 / 1/3
= 1

Therefore,
1) lim [ ( x^2+2x-15)/(x^2-9)] , x->3 = 4/3
2) lim { [ 2z]/ [ 4 + z )^1/2 - ( 4-z)^1/2]} , z->0 = 4
3) lim{[2-x^1/2]/[3-(2x+1)1/2]},x->4 = 3/2
4) lim{[x+27]/[(x)^1/3+3]}, x-> -27 = 1