We have $\lim_{n\to\infty} \frac{n^2}{2^{log(n)}} = \lim_{n\to\infty} \frac{2n}{2^{log(n)}log(2)(1/n)} = \lim_{n\to\infty} \frac{2n^2}{log(2) 2^{log(n)}}$ using L'Hospital's rule. But this process continues because although the power of $n$ in the numerator reduces as we differentiate, the denominator keeps providing the numerator with $n$ due to the differentiation of the $log(n)$ term.
Suppose we continue this process $\lceil{log(n)}\rceil$ times (does this make sense as $n\to \infty$ ?). Then we have $2^{log(n)}$ in the numerator, which cancels out with the $2^{log(n)}$ in the denominator. Does this mean that the limit tends to infinity? Does this mean in fact that $2^{log(n)}=O(n^2)$?