Using L'Hopital's rule, with
f(x) = x lnx= lnx /(1/x),
Limit as x->0 = Lim[(1/x)/(-1/x^2)]
= Lim(-x) = 0
There is no reason why the value of f(1) cannot also be the limit of f(x) as x-> 0
f(x) has a minimum between 0 and 1, at x = 1/e. The minimum is -0.3678
lim x-->0+ f(x), where f(x)= x ln x
values for x are 1,0.5,0.1,0.05,0.01,0.005,0.001
when i plugged in 1 i got 0 then -.34657 then -.23025 ................... leading up to -0.00690
what is the value of the limit is it approaching 0?? and if it is approaching 0 then why is the value for 1 --- 0 ?? :S
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