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.
Well,

$$\sqrt{x^6 - 9 x^3} = x^{3} \left(1 - \frac{9}{x^{3}}\right)^{1/2}$$

and so the question is really about understanding how $\sqrt{1 - t}$ looks when $t$ is pretty small. By noticing that the derivative $\frac{d}{dt} \sqrt{t} = \frac 1 2$ when $t = 1$, we can say that

$$\left(1 - \frac{9}{x^{3/2}}\right)^{1/2} \approx \frac 1 2 \left(\frac{-9} {x^{3}}\right)$$

which leads quickly to the claimed limit of $-9/2$.



Alternatively, there is a completely standard technique of multiplying and dividing by the conjugate, which is $\sqrt{x^6 - 9x^3} + x^3$, but I find that this isn't very enlightening.
T. Bongers' answer is right, but a bit sloppy.  A better way would be to use the binomial expansion.

$$\sqrt{1-9x^{-3}} = \sum_{n=0}^\infty \frac{(1/2)^{\color{red}{n}}(-9)^nx^{-3n}}{n!} = 1 - \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots$$
Where the red thing represents a falling power.
$$\lim_{x\to+\infty}\sqrt{x^6-9x^3}-x^3=\lim_{x\to+\infty}x^3(\sqrt{1-9x^{-3}} - 1) = \lim_{x\to+\infty}x^3\left(- \frac{9}{2x^{3}} - \frac{243}{8x^{6}} - \cdots\right) = -\frac{9}{2}$$