Is it possible to prove a slant asymptote using the limit of the derivative as it approaches infinity? If so, does the following work I provided count as proof for finding the equation instead of using the proper method of polynomial division to prove the asymptote? I'm not sure since this method was never mentioned in my Calculus course, but appears to work as a proof.
$$\text{Let $C$ be an arbitrary constant.}$$ $$\lim_{x\rightarrow\pm\infty}f'(x)=C$$
$$\text{Let $a$ be the $x$-value at which the tangent line was created.}$$ $$y=\lim_{a\rightarrow\pm\infty}\left[f'(a)(x-a)+f(a)\right]$$
I used the following function $f(x)$ to test if it worked: $$f(x)=\frac{-3x^3+2x}{2x^2}+4$$ $$\text{Slope of tangent line:}$$ $$m=\lim_{x\rightarrow\pm\infty}f'(x)=-\frac{3}{2}$$ $$\text{Final equation for asymptote:}$$ $$y=\lim_{a\rightarrow\pm\infty}\left[-\frac{3}{2}(x-a)+\left(\frac{-3a^3+2a}{2a^2}+4\right)\right]$$ $$y=-\frac{3}{2}x+4$$