Rewrite it as:

\(\displaystyle \frac{1}{6}\lim_{b\to {\infty}}\frac{a^{2}-ab}{-b-\sqrt{a^{2}-ab+b^{2}}}+\frac{a}{6}\)

Now, divide the top by b, the b in the bottom by b and the radical by \(\displaystyle \sqrt{b^{2}}\)

\(\displaystyle \frac{1}{6}\lim_{b\to {\infty}}\frac{\frac{a^{2}}{b}-\frac{ab}{b}}{\frac{-b}{b}-\sqrt{\frac{a^{2}}{b^{2}}-\frac{ab}{b^{2}}+\frac{b^{2}}{b^{2}}}}+\frac{a}{6}\)

Then, it whittles down to:

\(\displaystyle \frac{1}{6}\lim_{b\to {\infty}}\frac{-a}{-2}+\frac{a}{6}\)

\(\displaystyle \frac{a}{12}+\frac{a}{6}=\frac{a}{4}\)