<s>[INST] Why do I have to perform polynomial division when trying to find slant asymptotes When trying to find the slant asymptote of $\frac{2x^2+x}{x-3}$, the way I thought was correct is to divide everything by $x$ to get $\frac{2x+1}{1-\frac{3}x}$. All that was left was to say that as $x$ tends to $\infty$, $\frac{3}x$ tends to $0$, so the asymptote is $2x+1$. Spoiler: it was not.
If I would do it the polynomial division way, I would get that the asymptote is $2x+7$. My question is: what is wrong with my way?
Helpful link though it did not answer my question here
 [/INST] Asserting that $2x+1$ is an asymptote means that$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=0.$$But$$\frac{2x^2+x}{x-3}-(2x+1)=\frac{6x+3}{x-3}$$and therefore$$\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(2x+1)=6\ne0.$$On the other hand\begin{align}\lim_{x\to\infty}\frac{2x^2+x}{x-3}-(ax+b)=0&\iff\lim_{x\to\infty}\frac{-a x^2+2 x^2+3 a x-b x+x+3 b}{x-3}=0\\&\iff\left\{\begin{array}{l}-a+2=0\\3a-b+1=0\end{array}\right.\\&\iff\left\{\begin{array}{l}a=2\\b=7.\end{array}\right.\end{align}
</s>