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Answer 1 · 2016-12-15T17:04:00

#lim_(x->0) x^lnx =+oo#
#lim_(x->+oo) x^lnx =+oo#

Substitute:

#x=e^y#

So that:

#x^lnx= (e^y)^(lne^y)= (e^y)^y=e^(y^2)#

As #y=lnx#, we have that:

for #x->0#, #y->-oo#
for #x->+oo#, #y->+oo#

Now:

#lim_(x->0) x^lnx =lim_(y->-oo) e^(y^2) =+oo#
#lim_(x->+oo) x^lnx =lim_(y->+oo) e^(y^2) =+oo#

graph{x^lnx [-40, 40, -20, 20]}