The problem: why does $\displaystyle\lim_{x \rightarrow \infty} \dfrac{4x^3+6}{3x^3+\sqrt{36x^6+2}}$ NOT equal $\displaystyle\lim_{x \rightarrow -\infty} \dfrac{4x^3+6}{3x^3 + \sqrt{36x^6+2}}$?
A common (wrong) approach:
1. You notice that "$x^3$" is the highest power of $x$ in the problem (taking the $\sqrt{36x^6+2}$ into consideration)
2. You multiplied by the "convenient form of $1$", $\dfrac{\frac{1}{x^3}}{\frac{1}{x^3}}$ and got
$$\dfrac{4x^3+6}{3x^3+\sqrt{36x^6+2}} \dfrac{\frac{1}{x^3}}{\frac{1}{x^3}} = \dfrac{4+\frac{6}{x^3}}{3 + \sqrt{36+\frac{2}{x^6}}}$$
3. You then took the limit as $x \rightarrow -\infty$ and concluded that it should be $\dfrac{4}{3+6} = \dfrac{4}{9}$.
This method works for $x \rightarrow \infty$ but NOT when $x \rightarrow -\infty$.
This happens because: as $x \rightarrow -\infty$, we know
$4x^3 \rightarrow -\infty$ and
$3x^3 \rightarrow -\infty$, and
(the key part:) $\sqrt{36x^6} \rightarrow +\infty$
The fact that $\sqrt{36x^6}\rightarrow +\infty$ while $\sqrt{36x^6}=6x^3 \rightarrow -\infty$ is the issue. The first formula "erases the information" of the negative sign before multiplying by 36 and taking the root while the second formula "has forgotten" to erase the information about the negative and so the limit can tend to $-\infty$. If that's not a satisfying reason, there's more.
When you "multiply by $\dfrac{\frac{1}{x^3}}{\frac{1}{x^3}}$", you can't forget that you're working on the limit as $x \rightarrow -\infty$, and therefore $x^3$ is negative and so is $\dfrac{1}{x^3}$.
So the step where you pull the $\dfrac{1}{x^3}$ inside the square root in the bottom requires a lot of care! Just because you can't "see" the negative sign of $x$ doesn't mean it's not there. It may help you to rewrite, for $x<0$, $x^3=-|x^3|$ (but this is not strictly necessary).
So performing the step
$$\dfrac{1}{x^3}\sqrt{36x^6+2} = \sqrt{36+\frac{2}{x^6}}$$
is WRONG when $x$ is negative. It should look like
$$\dfrac{1}{x^3}\sqrt{36x^6+2} = -\sqrt{36+\frac{2}{x^6}}.$$
This will yield then the limit to be
$$\displaystyle\lim_{x \rightarrow -\infty} \dfrac{4}{2-\sqrt{36+\frac{2}{x^6}}}=\dfrac{4}{2-6} = -\dfrac{4}{3}.$$
