Question #79b6c

1 Answer
Dec 15, 2016

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

Explanation:

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]}