Limits at infinity concept Two examples that I'd like to discuss to help me clarify a key concept of limits at infinity.
Ex. 1  $\lim_{x\to \infty} (\sqrt{x^6+5}-x^3)$.  I argued (to myself) that as $x$ gets very large then the $5$ becomes negligible. So, I would get $\vert x^3\vert-x^3$.  For positive $x$ this $=0$.  Works.  Perfect.  Maybe this is a flawed concept and I just got lucky that it is zero.
Then I get to the example that I don't know why it doesn't work in the same way:
Ex. 2  $\lim_{x\to \infty} (\sqrt{x^6+5x^3}-x^3)$.  If I solve this by multiplying by the conjugate, I of course get the correct answer of $5/2.$  But it is not clear to me why the $5x^3$ term does not become negligible compared to the $x^6 $ term.  If it did, I'd get the same answer for Ex. 2 as I did for Ex. 1.
Can you assist me in understanding this? For now, my takeaway is to always use conjugate math when dealing with square roots.
 A: 
$\lim_{x\to \infty} (\sqrt{x^6+5}-x^3)$.  I argued (to myself) that as $x$ gets very large then the $5$ becomes negligible.

That works with division but not with subtraction, because with division the relative size of the $5$ by comparison to the size of the growing denominator goes to $0.$ Note that
$$
\lim_{x\to\infty} \Big((x+5) - x \Big) \ne 0
$$
although the $5$ similarly becomes negligible by comparison to either $x+5$ or $x.$
\begin{align}
& \sqrt{x^6+5} -x^3 \\[10pt]
= {} & \frac 5 {\sqrt{x^6+5} + x^3} \to 0.
\end{align}
This is a somewhat unenlightening example because of what it omits. Consider this similar-looking example:
\begin{align}
& \sqrt{x^6+5x^3} -x^3 \\[10pt]
= {} & \frac {5x^3} {\sqrt{x^6+5x^3} + x^3} \\[10pt]
= {} & \frac{5}{\sqrt{1 + \frac 5 {x^3}} + 1} \to \frac 5 2.
\end{align}
This example proves that you first approach was in error, since in this case the limit is not $0.$
(Note that if $x$ had been approaching $-\infty$ rather than $+\infty,$ then you would have $\sqrt{x^6} = |x^3| = -x^3$ and then the answer would look quite different.
