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sindhuja P Grade: 11
        

Lt      [1/x1/ln(e^x-1)+(1/x)sinx-xlogx]  (x>0,and [.] denotes


x→0                                                        greatest integer    


                                                                function)

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear paidepelly


solve each part individually


first part


Lt x→0     1/x1/ln(e^x-1)


 


eLt x→0 -(ln x)/ln(ex-1)


use L hospital rule and find limit


=1/e


 


for second part


Lt x→0(1/x)sinx


e Lt x→0sin x ln(1/x)


use L hospital rule and find limit 


  =1


for third part


Lt x→0-xlogx  =0


 Now add all three limit


  1/e +1+0


 =.367 +1+0 =1.367


so [1.367] =1 answer





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Badiuddin





7 years ago
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