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.

Hmmm, no one has pointed out the obvious:
$\lim \sqrt{x^6 - 9x^3} - x^3 = \lim \sqrt{x^6 - 9x^3 + 36/4} - x^3$
$ = \lim \sqrt{(x^3 - 9/2)^2} - x^3 = \lim x^3 - 9/2 - x^3 = -9/2$
Intuition.... hmm .... I guess realizing the $x^3$ from $\sqrt {x^6 + stuff}$ was going to cancel the $-x^3$.  So I want some $\lim \sqrt {Y_{x^3}^2} - x^3$ and figuring $Y_{x^3}^2$ must be.
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So why does $\lim \sqrt{x^6 -9x^3} = \lim\sqrt{x^6 -9x^3 + 36/4}$?
Let $\epsilon > 0$.  We wish to solve for which $x$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| = x^3 - \frac 92 - \sqrt{x^6-9x^3} < \epsilon$.  If we can show that this can be solved for all $x > M$ for some $M$ we are done.
$(x^3 - \frac 92) - \epsilon < \sqrt{x^6 - 9x^3} $.  For large enough $x$ we may assume this are both positive.
$x^6 - 9x^3 + 9 - 2\epsilon*(x^3 - \frac 92) + \epsilon^2 < x^6 - 9x^3$
$9 -  2\epsilon*(x^3 - \frac 92) + \epsilon^2 < 0$
$(x^3- \frac 92)  > \frac{(9 + \epsilon^2)}{2\epsilon}$
So for any $x > \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ this will be true.
So for all $\epsilon > 0$ if $x > M = \sqrt[3]{\frac 92 + \frac{(9 + \epsilon^2)}{2\epsilon}}$ we have $|\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}| < \epsilon$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}-\sqrt{x^6-9x^3}= 0$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4}= \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}$
So $\lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3}-x^3 = \lim_{x\rightarrow \infty}\sqrt{x^6 -9x^3 + 36/4} -x^3 = -\frac 92$
Or more generally...
If $c_x \rightarrow \infty$ and $f$ is continuous, then $\lim_{x\rightarrow \infty}f(c_x + h) = \lim_{x\rightarrow \infty}f(c_x(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x(\lim_{x_\rightarrow \infty}(1 + h/c_x))=\lim_{x\rightarrow \infty}f(c_x*1)=\lim_{x\rightarrow \infty}f(c_x)$
Let $c_x = x^6 - 9x$, $h= 9=36/4$ $f(y) = \sqrt{y}$.
