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

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


x→0                                                        greatest integer    


                                                                function)

Grade:11

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

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



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