$ \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...
 A: Multiply the fraction by $1$ in the carefully chosen disguise $\dfrac{e^{-x}}{e^{-x}}$ and do a bit of algebra:
$$\begin{align*}
\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}&=\lim_{x\to\infty}\left(\frac{xe^{x-1}}{(x-1)e^x}\cdot\frac{e^{-x}}{e^{-x}}\right)\\
&=\lim_{x\to\infty}\frac{xe^{-1}}{x-1}\\
&=\lim_{x\to\infty}\frac{x}{e(x-1)}\\
&=\frac1e\lim_{x\to\infty}\frac{x}{x-1}\\
&=\frac1e\lim_{x\to\infty}\frac{x-1+1}{x-1}\\
&=\frac1e\lim_{x\to\infty}\left(1+\frac1{x-1}\right)\;.
\end{align*}$$
That last limit really is easy.
You may wonder how I came up with some of the steps. The very first one was simply recognizing that if I divided numerator and denominator by $e^x$, the resulting fraction would be a lot simpler, in that all of the exponentials would be gone. Pulling constant factors outside the limit is usually useful and almost never hurts. The simplification of $\frac{x}{x-1}$ could also have been achieved by doing a straightforward polynomial long division of $x-1$ into $x$, but the trick of subtracting and adding the same amount (here $1$) in order to get an expression that can be split in some nice way is a pretty common one that’s worth remembering.
A: Clearly,
$$\frac{{x{e^{x - 1}}}}{{(x - 1){e^x}}} \sim \frac{{x{e^{x - 1}}}}{{x{e^x}}} = \frac{{{e^{x - 1}}}}{{{e^x}}}.$$
So ...
A: $$\lim_{x\to\infty}\frac{xe^{x-1}}{(x-1)e^x}=\lim_{x\to\infty}\frac{x}{x-1}\cdot\lim_{x\to\infty}\frac{e^{x-1}}{e^x}=1\cdot\frac{1}{e}=\frac{1}{e}$$
the first equality being justified by the fact that each of the right hand side limits exists finitely.
A: This is straightforward
$$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}=\lim_{x \to \infty} \frac{x}{(x-1)e}=\frac{1}{e}$$
(Chris)
