Question

Question #79b6c

Answer 1

`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]}