Answer
$-\frac{\pi}{2}$
Work Step by Step
Since $\lim\limits_{x \to \infty}\frac{x^{2}+9}{9-x}=\lim\limits_{x \to \infty}-\frac{x^{2}}{x}=\lim\limits_{x \to \infty}-x=-\infty$
And $\lim\limits_{t \to -\infty}tan^{-1}(t)=-\frac{\pi}{2}$
Then
$\lim\limits_{x \to \infty} tan^{-1}(\frac{x^{2}+9}{9-x}
)=-\frac{\pi}{2}$
