\begin{align*}
  \lim_{n\to\infty} a_n &= \lim_{n\to\infty} \left(\sqrt{n^2-n+9} - \sqrt{n^2+9}\right) \\[1ex]
  &= \lim_{n\to\infty} \left(\sqrt{n^2-n+9} - \sqrt{n^2+9}\right)
    \frac{\sqrt{n^2-n+9} + \sqrt{n^2+9}}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
  &= \lim_{n\to\infty} \frac{(n^2-n+9) - (n^2+9)}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
  &= \lim_{n\to\infty} \frac{-n}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
  &= -\lim_{n\to\infty} \frac{n}{\sqrt{n^2-n+9} + \sqrt{n^2+9}} \\[1ex]
  &= -\lim_{n\to\infty} \frac{1}{n^{-1}\sqrt{n^2-n+9} + n^{-1}\sqrt{n^2+9}} \\[1ex]
  &= -\lim_{n\to\infty} \frac{1}{\sqrt{n^{-2}(n^2-n+9)} + \sqrt{n^{-2}(n^2+9)}} \\[1ex]
    &= -\lim_{n\to\infty} \frac{1}{\sqrt{1-\frac{1}{n}+\frac{9}{n^2}} + \sqrt{1+\frac{9}{n^2}}} \\[1ex]
\end{align*}

Dermed blir 

\begin{align*}
  -\lim_{n\to\infty} \frac{1}{\sqrt{1-\frac{1}{n}+\frac{9}{n^2}} + \sqrt{1+\frac{9}{n^2}}} &= -\frac{1}{\sqrt{1-0+0} + \sqrt{1+0}} \\
  &= -\frac{1}{1 + 1} \\
  &= -\frac{1}{2} \\
\end{align*}
