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Lt [1/x1/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/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)
=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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