# lim x tending to 0 (((cos x)^(1/2) - (cos x)^(1/3)) / (sin x)^2).Please solve this.

Nandana
110 Points
6 years ago
hi !
Ans : -1/6
Limx--> 0 √cosx – (cosx)1/3     0  form [LHospital’s rule]
sin2 x              0

Limx--> 0  [ ½  (√cosx)½ – 1/3 (cosx)-2/3 ] sinx   = [ -½  (√cosx)-1 +1/3 (cosx)-2/3 ]
2 sin x cosx                                            2cosx
=  -1/2 +1/3 = -1/6
Debayan das
18 Points
6 years ago
Nandana
110 Points
6 years ago
hi !Sorry in my last answer I did a small mistake !Ans : -1/6 Limx--> 0 √cosx – (cosx)1/3 0 form [LHospital’s rule] sin2 x 0 Limx--> 0 [ ½ (√cosx)½ – 1/3 (cosx)-2/3 ] sinx = [ -½ (√cosx)-1 +1/3 (cosx)-2/3 ] 2 sin x cosx 2cosx Here is my mistake , = 1/2 (-1/2 +1/3) = -1/12 Now ok no ?It `s a calculation mistake , sorry & thank you very much for telling !
Nandana
110 Points
6 years ago
hi ,
sorry for my mistake !
actually, what I did is correct but in the last step I did forget to multiply by half
here is what I ‘m talking about

Limx--> 0 √cosx – (cosx)1/3     0  form [LHospital’s rule]
sin2 x              0

Limx--> 0  [ ½  (√cosx)½ – 1/3 (cosx)-2/3 ] sinx   = Limx--> 0 [ -½  (√cosx)-1 +1/3 (cosx)-2/3 ]
2 sin x cosx                                                   2cosx
=  ½ (1/3 -1/2) = -1/12