$ \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x} $ $$ \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x} $$
I don't know what to do. At all.
I've read the explanations in my book at least a thousand times, but they're over my head.
Oh, and I'm not allowed to use L'Hospital's rule. (I'm guessing it isn't needed for limits of this kind anyway. This one is supposedly simple - a beginners problem.) Most of the answers I've seen on the Internet simply says "use L'Hospital's rule".
Any help really appreciated. I'm so frustrated right now...

$$\lim_{x\to\infty}\frac{xe^{x-1}}{(x-1)e^x}=\lim_{x\to\infty}\frac{x}{x-1}\cdot\lim_{x\to\infty}\frac{e^{x-1}}{e^x}=1\cdot\frac{1}{e}=\frac{1}{e}$$
the first equality being justified by the fact that each of the right hand side limits exists finitely.
