<s>[INST] Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote 
$y = -2x - 3/2$.
How we get this asymptote?
 [/INST] Note that if $ax+b$ is an asymptote for the function $f(x)$ in $-\infty$, then$$a=\lim_{x\to -\infty}{f(x)\over x}\\b=\lim_{x\to -\infty}{f(x)-ax}$$therefore$$a{=\lim_{x\to -\infty}{\sqrt{x^3\over x-3}-x\over x}\\=\lim_{x\to -\infty}{-\sqrt{x\over x-3}-1}\\=-2}$$and $$b{=\lim_{x\to -\infty}\sqrt{x^3\over x-3}-x+2x\\=\lim_{x\to -\infty}\sqrt{x^3\over x-3}+x\\=\lim_{x\to -\infty}{{x^3\over x-3}-x^2\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{{3x^2\over x-3}\over \sqrt{x^3\over x-3}-x}\\=\lim_{x\to -\infty}{3x^2\over -\sqrt{x^3(x-3)}-x(x-3)}\\=\lim_{x\to -\infty}{3\over -\sqrt{x-3\over x}-{x\over x-3}}\\=-{3\over 2}}$$
</s>