I feel that $2^{\sqrt{\log{n}}}$ could be dramatically simplified, but I'm sure how. Aside from plugging in huge values to test the functions, any ideas on how I can prove this relationship?
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You can write $n^{-100}=e^{-100 \log n}$, and $2^{\sqrt{\log n}}=e^{\log 2 \sqrt{\log n}}$; does that help? (And are you missing a minus sign somewhere? Otherwise $n^{-100}\rightarrow 0$ and $2^{\sqrt{\log n}}\rightarrow \infty$ is sufficient.) – mjqxxxx Feb 10 '14 at 05:41