# what is value of $\lim_{n\to\infty} n^{-2}e^{(\log(n))^a}$?

What is value of $\displaystyle\lim_{n\to+\infty} n^{-2}e^{(\log(n))^a}$, where $a > 1$? I tried l'Hospital's rule but it gets complicated.

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## 2 Answers

$n^{-2}e^{(\log(n))^a} = e^{(\log(n))^a - 2 \log(n)}$. If $a$ is a constant greater than $1$ then $(\log(n))^a$ will dominate compared to $2 \log(n)$, so this expression will also go to infinity as $n$ goes to infinity.

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Thanks Wonder!! –  quartz Apr 17 '12 at 19:06

Hint: Consider taking the logarithm.

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Thanks Aryabhata!! –  quartz Apr 17 '12 at 19:06
@quartz: You are welcome! –  Aryabhata Apr 17 '12 at 20:29