I feel dumb for asking this, but I couldn't quite show that this limit is 0 (which I think is correct) whenever $a>0$:

$$\lim_{x\to\infty} xe^{-a\frac{x}{\ln x}}.$$

I tried using L'Hospital's rule which was mildly annoying given the amount of terms. I also couldn't figure out a good way to use comparison either.


Hint: try to use the fact that $\log x < \sqrt{x}$ for $x \gg 1$

  • $\begingroup$ Yep, thanks. I knew I was just not thinking of something obvious. $\endgroup$
    – Keaton
    Aug 28 '15 at 16:21

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