<s>[INST] Find the asymptotes of $xy^2 - y^2 - x^3 = 0$ I am lost with this question because there is no 3rd degree $y$ term and no 2nd degree $x$ term.
Can $y - x - 1/2 = 0$ and $y + x + 1/2 = 0$ be its asymptotes?
or $y - x - 1/3 = 0$ and $y + x + 1/2 = 0$
 [/INST] Now, $y=\sqrt{\frac{x^3}{x-1}}$. Apart from the vertical asymptote $x=1$, we have two oblique asymptotes (OAs) in the form $y=ax+b$. To find these let us make the following limit zero:
$\begin{align}
\lim_{x\rightarrow \pm\infty}\sqrt{\frac{x^3}{x-1}}-(ax+b)&=\lim_{x\rightarrow \pm\infty}\frac{\frac{x^3}{x-1}-(ax+b)^2}{\sqrt{\frac{x^3}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{x^2+x+1+\frac{1}{x-1}-a^2x^2-2abx-b^2}{\sqrt{x^2+x+1+\frac{1}{x-1}}+(ax+b)}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}+\frac{1}{x-1}}+ax+b}\\
&=\lim_{x\rightarrow \pm\infty}\frac{(1-a^2)x^2+(1-2ab)x+1-b^2+\frac{1}{x-1}}{|x+\frac{1}{2}|+ax+b}\\
\end{align}$
To make this limit zero at $\infty$: $a=1$ and $2b-1=0$. So, the OA at $\infty$ is $y=x+\frac{1}{2}.$
To make this limit zero at $-\infty$: $a=-1$ and $2b+1=0$. So, the OA at $-\infty$ is $y=-x-\frac{1}{2}.$
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