Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes). Source : Problem $4.2.$ , $n° 4 $
http://www.gycham.vd.ch/favre/2M/asymptotes/Asymptotes.pdf
The question I'm trying to solve is " determine, in case it exists, the equation of the horizontal asymptote of the following function"

$ \LARGE f(x)= \frac {4x+1} {\sqrt{x^2+9}}$

I do not understand the way Symbolab applies the rule " divide by the highest power of the denominator" to this question.
According to Symbolab, the result of applying this rule yields :

$ \LARGE f(x)= \frac {4+ \frac {1}{x}} {\sqrt {1+\frac {9} {x^2}}}$

This raises  some questions :
(1) what is supposed to be the power of the denominator, in case the denominator is a polynomial placed under a radical sign? is there an official rule regarding this point?
(2) how does the transformed expression result from dividing by the same quantity both the numerator and the denominator? the numerator has been dvided by $x$, is it also the case of the denominator?

 A: 
what is supposed to be the power of the denominator, in case the denominator is a polynomial placed under a radical sign?

Informally, we can think of this by reasoning "if $x$ is very large, then $9$ is insignificant compared to $x^2$, so $x^2+9$ acts like $x^2$ and $\sqrt{x^2+9}$ acts like $\sqrt{x^2}=|x|$. This doesn't prove anything in particular, especially since I haven't clarified exactly what "acts like" means, but it justifies the idea of dividing numerator and denominator by $|x|$.

how does the transformed expression result from dividing by the same quantity both the numerator and the denominator? the numerator has been dvided by $x$, is it also the case of the denominator?

Yes, dividing any fraction's numerator and denominator by any non-zero number (or formula resulting in a number) does not change the value of the fraction:
$$ \frac{A / R}{B / R} = \frac{A}{B} \cdot \frac{1/R}{1/R} = \frac{A}{B} \cdot 1 = \frac{A}{B} $$
But there is something strange in the result
$$ f(x) = \frac{4 + \frac{1}{x}}{\sqrt{1+\frac{9}{x^2}}} $$
-- it's correct when $x>0$ but incorrect when $x<0$. Since $\sqrt{x^2}=|x|$, we need to divide numerator and denominator by $|x|$, not by $x$:
$$ \begin{align*}
 f(x) &= \frac{\frac{4x+1}{|x|}}{\frac{1}{|x|}\sqrt{x^2+9}} \\
 f(x) &= \frac{4 \frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{\frac{1}{x^2}}\sqrt{x^2+9}} \\
 f(x) &= \frac{4 \frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{1+\frac{9}{x^2}}}
\end{align*} $$
When $x>0$, $|x|=x$ and this gives the formula you quoted. When $x<0$, $|x|=-x$ and it instead gives
$$ f(x) = \frac{-4 - \frac{1}{x}}{\sqrt{1+\frac{9}{x^2}}} $$
So the graph approaches two different horizontal asymptotes, at $y=+4$ and $y=-4$:
$$ \lim_{x \to +\infty} f(x) = 4 $$
$$ \lim_{x \to -\infty} f(x) = -4 $$
A: As you noted, $\space x^1\space $ is the highest power in $\space \sqrt{x^2+9}\space $ because, for large values of $\space  x, \space $ the $\space 9\space$ becomes insignificant.
Dividing both numerator and denominator by this yields
$ f(x)= \dfrac {4+ \dfrac {1}{x}} {\sqrt {1+\dfrac {9} {x^2}}}.\quad$
Now we can see that
$\space \dfrac{1}{x^n},n\in\mathbb{N} \space $ approaches
zero as $\space  x \space $ approaches infinity. The same applies if $\space  x \space $ is a polynomial that approaches infinity.
In any case, here, the $\space x$-terms disappear leaving $\quad f(x)=y=\pm 4\space$ if you consider both the positive and negative roots.
A: For large positive $x$ we have
$$f(x)=\sqrt{\frac{(4x+1)^2}{x^2+9}}
=\sqrt{\frac{16x^2+8x+1}{x^2+9}}
=\sqrt{\frac{16+8/x+1/x^2}{1+9/x^2}}.$$
For large negative $x$ we get
$$f(x)=-\sqrt{\frac{16+8/x+1/x^2}{1+9/x^2}}.$$
A: Informally, if you're going to let $n$ approach infinity, then all but the highest power of a polynomial can be ignored. In your example, as $x$ approaches infinity
$\sqrt{x^2+9} \rightarrow \sqrt{x^2} \rightarrow x $
and $\sqrt{(4x+1)^2} = 4x+1 \rightarrow 4x$
So $\sqrt{\frac{(4x+1)^2}{x^2+9}} \rightarrow \frac{4x}{x} \rightarrow  4$
Basically, you ignore all but the variables with the largest exponents and simplify.
Formally, you see that $x$ is the variable with the highest exponent in both the numerator and the denominator. So you divide both numerator and denominator by $x$, as you did.
