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$$ \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x} $$

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...

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What is $e^{x-1}/e^x$? – wj32 Oct 18 '12 at 2:50
Do you know any of the limit laws? Such as $\lim cf = c\lim f$? They are quite helpful in this case. – Raymond Cheng Oct 18 '12 at 2:53
up vote 4 down vote accepted

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.

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$$\frac{{x{e^{x - 1}}}}{{(x - 1){e^x}}} \sim \frac{{x{e^{x - 1}}}}{{x{e^x}}} = \frac{{{e^{x - 1}}}}{{{e^x}}}.$$

So ...

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the first equality being justified by the fact that each of the right hand side limits exists finitely.

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first downvote didn't come from me. I usually don't downvote neither questions, nor answers. – user 1618033 Oct 20 '12 at 5:42
THere seems to be a serial downvoter around here. I could almost have sworn it was you but, of course, I didn't since I'm not sure. I don't care either, to be honest, as in any case the serial downvotes are reversed back by the system and also because 2,4, or 40 points less or more make no real difference. – DonAntonio Oct 20 '12 at 17:20
honestly, if I wanted to downvote you then I'd have no problem telling you this direclty. I didn't downvote you this time. I downvote you ONLY when you downvote me. I do not like to downvote people (neither do I, nor my sister). – user 1618033 Oct 20 '12 at 17:25
then after I saw that you downvoted me I downvoted you. I mean that the first downvote didn't come from me. – user 1618033 Oct 20 '12 at 17:27
As before, I won't get into long discussions with you and I'll end this one after this message: I've no idea what you're talking about. I didn't downvote you, and I really don't care whether you downvoted me because you thought I did with you or because any other reason. I'm not interested in this kids game and please do feel as free as you want to downvote me: I don't care. Have a good day. – DonAntonio Oct 20 '12 at 18:25

This is straightforward

$$\lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}=\lim_{x \to \infty} \frac{x}{(x-1)e}=\frac{1}{e}$$


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