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.
what is value of $\lim_{n\to\infty} n^{-2}e^{(\log(n))^a}$?
6
$\begingroup$
real-analysis
limits
2 Answers
6
$n^{-2}e^{(\log(n))^a} = e^{(\log(n))^a - 2 \log(n)}$. 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.
6
Hint: Consider taking the logarithm.
-
0@quartz: You are welcome! – 2012-04-17