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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.

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2 Answers 2

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As you want only a hint:

You may set $n=e^{t^2}$ and now consider the new expression for $t \to +\infty$.

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  • $\begingroup$ How did you know to do that? $\endgroup$
    – eje211
    Feb 6, 2020 at 3:49
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    $\begingroup$ @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. $\endgroup$
    – trancelocation
    Feb 6, 2020 at 3:51
  • $\begingroup$ 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! $\endgroup$
    – eje211
    Feb 6, 2020 at 3:52
  • $\begingroup$ @eje211 : This trick is quite often useful. So, good to keep this in mind. Using this trick, you can often simplify seemingly ugly expressions into much nicer ones. $\endgroup$
    – trancelocation
    Feb 6, 2020 at 3:53
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    $\begingroup$ @eje211 : If you look around here on this platform, you will see that "intuiting" a nice substitution can have a tremendous simplifying character to solving a problem. Just check out some answers to limits, integrals or inequalities. To somehow have a good gut feeling which substitution may work, I find this definitely also to be a creative process. Of course, not always but quite often. $\endgroup$
    – trancelocation
    Feb 6, 2020 at 4:23
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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.

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