Divide by the highest power or divide by the denominator highest power? When resolving limits that $x\to \pm ∞$ the teacher taught us to divide by the highest power. But I've seen some that divide by the highest power in the denominator
First thing it comes to my mind is that it can be archived either way.
if I divide by the highest power in the entire fraction
$$\Large\lim _{x\to \infty \:}\left(\frac{4x^2+x^6}{1-5x^3}\right)\:$$
$$=\Large\lim \:_{x\to \:\infty \:\:}=\left(\frac{\frac{4x^2}{x^6}+\frac{x^6}{x^6}}{\frac{1}{x^6}-\frac{5x^3}{x^6}}\right)$$
$$=\Large\lim \:_{x\to \:\infty \:\:}=\left(\frac{\frac{4}{x^4}+1}{\frac{1}{x^6}-\frac{5}{x^3}}\right)=\:\frac{0+1}{0-0}\:=\:\frac{1}{0}\:=\:∞?$$
But... continuing with the same example above, If I divide by the highest power in the denominator:
$$\Large\lim _{x\to \infty \:}\left(\frac{4x^2+x^6}{1-5x^3}\right)\:$$
$$\Large=\lim \:_{x\to \:\infty \:\:}\left(\frac{\frac{4x^2}{x^3}+\frac{x^6}{x^3}}{\frac{1}{x^3}-\frac{5x^3}{x^3}}\right)\:$$
$$\Large=\lim \:\:_{x\to \:\:\infty \:\:\:}\left(\frac{\frac{4}{x}+x^3}{\frac{1}{x^3}-5}\right)\:=\frac{∞}{-5}\:=\:-∞$$
My questions are:

*

*Can it be archived both ways? If so what I'm doing wrong in the first example?

*If the method you use depends on something, on what?

 A: So, at one stage, you have
$$L = \lim_{x \to \infty} \frac{1 + 4/x^4}{1/x^6 - 5/x^3}$$
Firstly, $1/0$ does not mean anything. It is not $\infty,-\infty,$ or anything; it is simply undefined. Hence, taking the limit as you did and getting $1/0$ as an answer (regardless of the fact you conclude $L = \infty$ thereafter) is a hint you are probably doing something wrong, or at least overlooking a nuance of the situation.
Consider the fact that $x^3 < x^6$ for $x$ large (and in particular as $x \to \infty$). Then $1/x^6 < 1/x^3 < 5/x^3$, right? And then therefore,
$$\frac{1}{x^6} - \frac{5}{x^3} < 0$$
Think of the ramifications of this as you take the limit: the numerator is growing closer and closer to $1$, but the denominator is growing closer and closer to zero from the left (negative) side of $0$. Whatever that denominator may be can be incredibly small, but it will be negative. Hence,
$$L = \lim_{x \to 0^-} \frac 1 x = -\infty$$
would be the more correct and formal claim. (You intuited that the limit was $1/x$, but did not consider the direction and went with a common "default" of $x \to 0^+$ giving $+\infty$.)
A: 
"...the purpose is to bring the denominator to a form with a finite non-zero limit..." -dxiv

I think this is the key insight toward making these problems easier; and I'll take it a step further saying that you want to do it for the numerator as well!

So what is the problem at hand?
$$
\lim _{x\to \infty\:}\left(\frac{4x^2+x^6}{1-5x^3}\right)=\ ?
$$
So, we have a polynomial of order $x^6$ in the numerator, and a polynomial of order $x^3$ in the denominator... so what wins out in the limit?
Following dxiv's advice about the denominator, we can rearrange that a bit by dividing out powers of $x$ until all that remains should be finite when we take the limit... in other words:
$$
\lim _{x\to \infty \:} \left(1-5x^3\right) = \lim _{x\to \infty \:} x^3 \left(-5 + \frac{1}{x^3} \right)
$$
The $x^3$ outside will still need to be contended with later, but we do at least know that the part left inside the parentheses will stay finite as we approach that infinite limit.
If we play the same games with the numerator:
$$
\lim _{x\to \infty \:} \left(4x^2+x^6\right) = \lim _{x\to \infty \:} x^6 \left(1 + \frac{4}{x^4} \right)
$$
then we once again have an exterior infinity to deal with, but the portion inside will stay finite as we approach the infinite limit.
And then, when we finally bring it all together:
$$
\lim _{x\to \infty\:}\left(\frac{4x^2+x^6}{1-5x^3}\right)=
\lim _{x\to \infty\:}\left(\frac{x^6 (1+\frac{4}{x^4})}{x^3 (-5+\frac{1}{x^3})} \right) =\:?
$$
then we see that we have two finite quantities in parentheses, and outside that we have some $x$ powers that will partially cancel out.
$$
\lim _{x\to \infty\:}\left(\frac{4x^2+x^6}{1-5x^3}\right)=
\lim _{x\to \infty\:}\left(x^3 \frac{(1+\frac{4}{x^4})}{(-5+\frac{1}{x^3})} \right) \approx \lim _{x\to \infty\:} \frac{x^3}{-5}
$$
A: These are both heuristic suggestions.  They work enough of the time to try them, and there is no reason except lack of success to stop trying one of them.
If you distinguish between $+\infty$ and $-\infty$ note that you need to consider the signs.  Are your limits $x \to +\infty$?  Or can $x$ get large in either direction.  In your first example the denominator has the sign of $x$ so the limit as $x \to -\infty$ is $-\infty$.  In your second example the numerator has the sign of $x$ while the denominator is negative, so the limit as $x \to -\infty$ is $+\infty$.  You need to be consistent.
