Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes).

Solution 1:

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 $y=-4$:

$$ \lim_{x \to +\infty} f(x) = 4 $$ $$ \lim_{x \to -\infty} f(x) = -4 $$

Solution 2:

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.

Solution 3:

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 $$f(x)=-\sqrt{\frac{16+8/x+1/x^2}{1+9/x^2}}.$$