Why $\frac{n^3}{2^{\Omega(\sqrt{\log n})}}$ doesn't refute the lower bound $O(n^{3-\delta})$?
It looks like the comparison in the question is wrong.

$$\begin{aligned}
\lim_{n -&gt; \infty} \frac{\frac{n^3}{2^{\sqrt{\log n}}}}{\ \ n^{3-\delta}\ \ } &amp;= \lim_{n -&gt; \infty} \frac{n^{\delta} }{2^{\sqrt{\log n}}}=\lim_{n -&gt; \infty} \frac{2^{\delta\log n} }{2^{\sqrt{\log n}}}\\
&amp;=\lim_{n -&gt; \infty} 2^{\delta\log n-\sqrt{\log n}}=\lim_{m -&gt; \infty}2^{\delta m-\sqrt{m}}\\
&amp;=\lim_{m -&gt; \infty}2^{\sqrt m(\delta\sqrt{m} -1)}\\
&amp;=2^{+\infty}=+\infty.\\
\end{aligned}$$

Here $\log n$ is understood as $\log_2n$. Since $\log_a n=\log_a2\cdot\log_2n$, the limit above will still be infinity if the base of $\log$ is switched to any number greater than 1.

In fact, for any constant $c&gt;0$, however smaller it is and any constant $\delta&gt;0$, however small it is, we still have, by similar argument,

$$\lim_{n \to \infty}\frac{\frac{n^3}{2^{c\sqrt{\log n}}}}{\ \ \ {n^{3-\delta}\ \ \ }} =+\infty.$$




  I compare between $\lim_{n -&gt; \infty} \frac{ \frac{n^3}{2^{\sqrt{\log n}}}}{\ \ \ n^{3-\delta}\ \ } = 0 $ So, $\frac{n^3}{2^{\Omega(\sqrt{\log n})}}$ is better than the other one, so why it doesn't mean that we refute the conjecture.


To be more careful, there is another mistake in the above reasoning. Even if $\lim_{n -&gt; \infty} \dfrac{\frac{n^3}{2^{\sqrt{\log n}}}}{\ \ n^{3-\delta}\ \ } = 0 $, it does not imply it will refute the conjecture. The point is, "APSP can be solved in time $\dfrac{n^3}{2^{\Omega(\sqrt{\log n\ })}}$" might be true because it can be solved in time $\frac{n^3}{2^{0.01\sqrt{\log n\ \ }}}$, not because it can be solved in time $\frac{n^3}{2^{\sqrt{\log n\ \ }}}$". If you want to use the fact that "APSP can be solved in time $\dfrac{n^3}{2^{\Omega(\sqrt{\log n\ })}}$" to refute the conjecture, you have to show that 

$$O(n^{3-\delta}) \cap \frac{n^3}{2^{\Omega(\sqrt{\log n})}} = \emptyset,$$
or, in plain words, there is no function $f\in \Omega(\sqrt{\log n})$ such that $\dfrac{n^3}f\in O(n^{3-\delta})$. 

Well, in fact, the other extreme is true. $\dfrac{n^3}f\in O(n^{3-\delta})$ for all $f\in \Omega(\sqrt{\log n})$.