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 a fixed positive $k$ we should expect $\sqrt {y+k}$ to be very close to $\sqrt y$ when $y$ is very large, because  when $d$ is not  close to $0,$ we expect a big difference between $(d+\sqrt y)^2$ and $y,$ which  means $d+\sqrt y$ is too big to be $\sqrt {y+k}$.
We  can confirm this  by observing that if $k,y$ are positive and $\sqrt {y+k}=\sqrt y+e_y$ then $$y+k=(\sqrt y +e_y)^2=y +2e_y\sqrt y +e_y^2>y+2e_y\sqrt y$$ which implies $0<e_y<k/2\sqrt y.$
With $y=x^6-9x^3$ and $k=(9/2)^3,$ we have $x^3-9/2=\sqrt {y+k}=e_y+\sqrt y.$
