1. Improper integral

I need some help with the limit. I remember l'hospital's rule, but when I apply it; it really becomes a mess. I can't help thinking that I'm over thinking this... though it may just be as arduous as I think it is.

edit: ignore the minus sign after [(e^(1/x))/x], rushed job in equation editor

2. Originally Posted by ZeroAxiom

I need some help with the limit. I remember l'hospital's rule, but when I apply it; it really becomes a mess. I can't help thinking that I'm over thinking this... though it may just be as arduous as I think it is.

edit: ignore the minus sign after [(e^(1/x))/x], rushed job in equation editor
This limit is $+\infty$. Another way to see this is to procede to the change of variable $u=\frac{1}{x}$: the integral becomes $\int_0^1 \frac{e^{1/x}}{x}\frac{dx}{x^2}=\int_1^\infty u e^u du=+\infty$, obviously (the function in the integral goes to $+\infty$ when $u\to\infty$).