# Power of a square root of a logarithm

I'll be frank. This is homework. And I'm not actually asking for the answer. Just for... something.

$$\frac{f(n)}{g(n)} = \frac{n^a}{2^{\sqrt{\ln n}}}, a \in \mathbf{N}, a > 1$$

The point is to apply a limit and get to either a constant, zero or infinity, when $$n$$ gets close to infinity, obviously. It seems that I need to move the $$n$$ in the denominator to the numerator, but it's just dug in so deep. This is a fraction, so the only operation I can do on both sides is multiplication and there are no common factors on both sides. I can do a Lhopital rule, but deriving the denominator will still get me a variation of the original value.

Basically, is there any hint on how I can get rid of the $$2$$ and just reach the square root of the natural logarithm directly? Then, things are obviously much easier. Or change the denominator so that the derivation produces something useful?

Again, I'm not looking for the solution, just for a starting point. I tried many things and I never seem to be getting anywhere.

You may set $$n=e^{t^2}$$ and now consider the new expression for $$t \to +\infty$$.
• @eje211 : In order to get rid of the logarithm, I need $e^{something}$. To get rid of the root, that something should be a square. Feb 6, 2020 at 3:51
• Oh, because then, I have an $e$ where my $n$ was, but also a variable and a square! I never would have thought of changing the variable! Feb 6, 2020 at 3:52
Write $$k=\sqrt{\ln n}$$. We have $$k\to\infty$$ when $$n\to\infty$$ and $$n=e^{k^2}$$. $$\frac{n^a}{2^{\sqrt{\ln n}}}=\frac{e^{ak^2}}{2^k}$$ Now something should be clear.