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

Debayan das , 7 Years ago
Grade 11
anser 4 Answers
Nandana

Last Activity: 7 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

Last Activity: 7 Years ago

The answer of this question is -1/12.Your answer is not correct .The answer of this question is -1/12.Your answer is not correct .

Nandana

Last Activity: 7 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

Last Activity: 7 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
                                as your Answer !
    

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