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Today I was asked if you can determine the divergence of $$\int_0^\infty \frac{e^x}{x}dx$$ using the limit comparison test.

I've tried things like $e^x$, $\frac{1}{x}$, I even tried changing bounds by picking $x=\ln u$, then $dx=\frac{1}{u}du$. Then the integral, with bounds changed becomes $\int_1^\infty \frac{1}{\ln u}du$ This didn't help either.

This problem intrigued me, so any helpful pointers would be greatly appreciated.

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    $\frac{e^x}{x} \geq \frac{1}{x}$ when $x >0$.2012-07-17
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    that's good for direct comparison, which I know can be done easily with this. But I wanted to do it with limit comparison2012-07-17
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    Do you mean $e^{x}$ or $e^{-x}$? If it's $e^x$, you know that $e^x/x\rightarrow\infty$ as $x\to\infty$ and you are dead right there.2012-07-17
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    @ncmathsadist no, it is $e^x$, I know this function is divergent, it is a matter of showing it using a certain method2012-07-18

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