Considering expressions such as $$y=\frac{3x^3 + 5x - 2}{2x^2 + 1}$$ you could use long division and get $$y=\frac{3 x}{2}+\frac{7}{4 x}+\cdots$$ which shows the slant asymptote and how the function approaches it.
Otherwise, you could factor the highest degree in both numerator and denominator to write $$y=\frac{x^3\left(3+\frac 5 x-\frac 2{x^3}\right)}{x^2\left(2+\frac 1{x^2}\right)}=x\frac{\left(3+\frac 5 x-\frac 2{x^3}\right)}{\left(2+\frac 1{x^2}\right)}$$ Since $x\to \infty$, all terms with $x$ in denominator becomes negligible and you just keep the constants.
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 $y=-4$:
$$ \lim_{x \to +\infty} f(x) = 4 $$
$$ \lim_{x \to -\infty} f(x) = -4 $$
Best Answer
Because terms of the form $\frac{1}{x^n}$ converge to zero as $x \to \infty$.
In your example above, $ \lim_{x \to \infty}\frac{10x^3}{2x^3+3x^2+6x} = \lim_{x \to \infty}\frac{10}{2+3\frac{1}{x}+6\frac{1}{x^2}} = \frac{10}{2} = 5$.