I need to prove that $\frac{6x-3-2x^3}{4x\:-\:2}\:$ is $\theta \cdot \left(x^2\right)$.
They way I tried doing this was:
At first, I tried solving for $O({x^2})$
$$\frac{\left(6x-3-2x^3\right)}{4x\:-\:2}\:\le \frac{6x}{4x}\ \le \frac{6}{4} \le \frac{6x^2}{4} $$
Then I tried solving for $\Omega(x^2)$:
$$\frac{\left(6x-3-2x^3\right)}{4x\:-\:2}\:\ge \frac{\left(6x^3-3x^3-2x^3\right)}{4x\:-\:2}\: \ge \frac{x^3}{4x} \le \frac{1x^2}{4} $$
I don't think I did this in a right way. I'm kinda lost.
If you want to prove $f(n)=\theta (g(n))$, prove that $\lim_{n \to \infty} \frac{f(n)}{g(n)}$ is a constant.
Here $$\lim_{x \to \infty}\frac{\frac{6x-3-2x^3}{4x-2}}{x^2}=-1/2$$ So $f(x)=\theta(g(x))$