You might be making this more complicated than necessary!
As \(x\rightarrow -\infty\) the 2 in the numerator is negligible compared with 9x2, and the -9 in the denominator is negligible compared with the 2x. So:
\(\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2+2}}{2x-9}\rightarrow\lim_{x\rightarrow -\infty}\frac{\sqrt{9x^2}}{2x}\rightarrow \lim_{x\rightarrow -\infty}\frac{3|x|}{2x}\)
Now, when x is negative \(\frac{|x|}{x}=-1\) so the limit is -3/2.