# Limit of $\frac{n^{1/2}}{5^{\log n}}$

I have to show that the limit of $\dfrac{n^{1/2}}{5^{\log n}}$ is equal to as $n$ goes to infinity.

I have seen in some cases people using logarithms on the numerator and the denominator and say that the original fraction goes to infinity(or zero) if the logarithms go to infinity (or zero). Is this valid? And when it does not work? For example, in this case if i work with the logarithms it comes out to be a non zero constant, so this approach is wrong. L'hospital on the other hand doesn't seem practical. Any ideas?

• Hint: $a^{\log b}=b^{\log a}$. – Lucian Nov 2 '14 at 12:42

$$\log \dfrac{n^{1/2}}{5^{\log n}} = \frac{1}{2}\log n -\log 5 \log n \to -\infty$$
since $\frac{1}{2}< \log5$, then use the continuity of $\log(x)$ to conclude that $$\dfrac{n^{1/2}}{5^{\log n}} \to 0$$
$\dfrac{\log \frac{1}{n}}{ \log \frac{1}{n^2}} \to \frac{1}{2}$, so $\dfrac{\frac{1}{n}}{\frac{1}{n^2}} \to e^{\frac{1}{2}}$, which is obviously wrong.
$$\lim_{n \rightarrow \infty}{\frac{n^{1/2}}{5^{\log(n)}}} = \exp(\lim_{n \rightarrow \infty}{\log(\frac{n^{1/2}}{5^{\log(n)}}})) = \exp(\lim_{n \rightarrow \infty}\left({\frac{1}{2}\log(n) - \log(5)\log(n)}\right)) = 0$$ since $\frac{1}{2} - \log(5) < 0$.