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I need help to calculate this limit $$\large\lim_{n\to \infty} n^{\frac {\log n}{n^2}}.$$

I know the limit of the exponent goes to zero, but the limit of the base goes to infinity, I think the limit goes to 1, am I right?

Thanks a lot

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Do you know how to use logarithms to simplify this kind of limit? – Álvaro Lozano-Robledo Feb 21 '13 at 3:42
up vote 7 down vote accepted

Hint: Expanding on Andre's hint:

$$\log\left(n^{\frac{\log n}{n^2}}\right)=\frac{\log n\cdot\log n}{n^2}=\left(\frac{\log n}{n}\right)^2$$

Recall that by the continuity of $\log$ we have that $\lim_{n\to\infty}\log(f(n)) = \log(\lim_{n\to\infty} f(n))$.

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thank you for your answer – user42912 Feb 21 '13 at 4:02
@Clayton: Yes, thanks! :-) – Asaf Karagila Feb 21 '13 at 4:06

Hint: Take the logarithm, using $\log(a^b)=b\log a$. The result will the square of something you undoubtedly know about.

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I've already tried this, without success – user42912 Feb 21 '13 at 3:46
The logarithm is $\frac{\log n}{n}\log n=\left(\frac{\log n}{n}\right)^2$. I think you know that $\frac{\log n}{n}$ approaches $0$ as $n\to \infty$. So our limit is $e^0$. – André Nicolas Feb 21 '13 at 3:50
Thank you for your answer – user42912 Feb 21 '13 at 4:02

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