Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything simpler. Also tried expressing $x^3$ as $\sqrt{x^6}$, but that doesn't really help either.
One would think that, as $x^6$ grows more quickly than $x^3$ by a factor of $x^3$, the contribution of the $x^3$ term to the term in the square root would be dwarfed by the contribution of the the $x^6$ term, so the overall behavior of the first term in the limit would "behave" like $x^3$, as x gets bigger and bigger, so I would think intuitively that the limit would evaluate to 0.

For intuition, first get rid of the annoying cubes,
$$\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3=\lim_{x\to\infty}\sqrt{x^2 - 9x}-x.$$
Then the classical trick is to multiply/divide by the conjugate binomial and simplify,
$$\lim_{x\to\infty}\left(\sqrt{x^2 - 9x}-x\right)\frac{\sqrt{x^2 - 9x}+x}{\sqrt{x^2 - 9x}+x}=\lim_{x\to\infty}\frac{-9x}{\sqrt{x^2 - 9x}+x}=-\frac92.$$
Another approach is to factor out $x$ and use the Taylor development $\sqrt{1+t}\approx 1+t/2$:
$$\lim_{x\to\infty}x\left(\sqrt{1 - \frac9x}-1\right)=\lim_{x\to\infty}x\left(- \frac9{2x}\right).$$

You can "feel" why the answer is not $0$ by comparing the plots of the functions $\sqrt{x^2-9x}$ and $x$.

