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

x→0 greatest integer

function)

Dear paidepelly

solve each part individually

first part

Lt _x→0     1/x^{1/ln(e^x-1)}

e^{Lt _x→0 -(ln x)/}ln(e^x-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