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I had this problem on my math test, and was stuck on it for quite some time.

$\lim_{x \to \infty}x^e/e^x$

I knew that the bottom grew faster than the top, but I didn't know how to prove it. I wrote that the limit approaches 0, but I am not sure how to prove it mathematically.

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    $\begingroup$ From the Taylor series, $$e^x \ge \frac{x^4}{4!}.$$ $\endgroup$
    – user296602
    Oct 11, 2017 at 3:42
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    $\begingroup$ Have you learnt L'Hospital's rule? $\endgroup$
    – MonkeyKing
    Oct 11, 2017 at 3:42
  • $\begingroup$ @MonkeyKing yes, I kept getting infinty over infinty $\endgroup$
    – Sergei Levashov
    Oct 11, 2017 at 3:43

2 Answers 2

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Show first that it is in indeterminate form.

Then perform L'Hopital's rule, differentiating the top and bottom.

$$\lim_{x \to \infty} \frac{x^e}{e^x}= \lim_{x \to \infty} \frac{ex^{e-1}}{e^x}=e(e-1)(e-2)\lim_{x \to \infty} \frac{1}{x^{3-e}e^x}=0$$

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  • $\begingroup$ I did that and kept getting infinity over infinity $\endgroup$
    – Sergei Levashov
    Oct 11, 2017 at 3:43
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    $\begingroup$ @SergeiLevashov From the moment the exponent in the numerator becomes less than one, it's time to stop... $\endgroup$
    – imranfat
    Oct 11, 2017 at 3:44
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If $x>0$ then $e^x>\frac{x^3}{6}$, by the power series for $e^x$.

So $\frac{x^e}{e^x}<\frac{6}{x^{3-e}}\to 0$ as $x\to\infty$.

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