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I don't know how to find the limit $$\lim_{x\to\infty}\frac{e^x}{\log x}.$$ How can I do this ?

Thank you in advance.

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    $\begingroup$ This is highly dependent on what you can use. $\endgroup$ – Git Gud Mar 10 '14 at 17:02
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For $x > 1$, $$ \frac{e^x}{\ln x} > \frac{1 + x + \frac{x^2}{2!}}{x} > \frac{x}{2} \to \infty. $$ The relevant inequalities can be proved by elementary means; see for example here and here.

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  • $\begingroup$ @DanielFischer I see. Thanks :) $\endgroup$ – 6005 Mar 10 '14 at 17:18
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$\frac{\exp(x)}{\ln(x)} > \frac{\exp(x)}{x} > x$

First part because $x > \ln(x)$

Second part because $\exp(x) > x^2$ as $x^2 = \exp(2 \cdot \ln(x))$ and $x > 2 \cdot \ln(x)$ for $x$ greater than 2. And exponential is an increasing function

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  • $\begingroup$ Hi, you should use \exp for the exponential function, \ln for natural log, and \cdot for multiplication. $\endgroup$ – 6005 Mar 10 '14 at 17:25
  • $\begingroup$ I see that you edited with the correct format, but I can't see the edited post for some reason. I'll make sure to remember this. $\endgroup$ – T_O Mar 10 '14 at 17:34
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You may also use the L'Hôpital's rule, see http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule which, in this case, in view of the answer given by Goos, would be like killing a fly with a cannon...

I am giving you this information because L'Hôpital's rule is quite useful in some much more difficult problems involving limits (see the link for such examples), so it's good to know it.

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$$e^x>x\log x,\quad x\ge 1,$$ since $e=e^1<1\cdot\log 1=0$ and $$e^x=\frac{d}{dx}e^x<\frac{d}{dx}(x\log x)=1+\log x.$$

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