What are the asymptote(s) and hole(s), if any, of  f(x) =(sqrt(3x)/(x-4))^3?

Mar 10, 2016

Asymptotes are $x = 4 \text{ and } y = 0$

Explanation:

Write as:$\text{ } f \left(x\right) = \frac{3 x \sqrt{3 x}}{{\left(x - 4\right)}^{3}}$

$\textcolor{b l u e}{\text{Point 1}}$

We know that the function will be undefined as the denominator approaches zero.

Thus ${\lim}_{x \to 4} f \left(x\right) = {\lim}_{x \to 4} \frac{3 x \sqrt{3 x}}{{\left(x - 4\right)}^{3}} \to \frac{12 \sqrt{12}}{0}$

$\textcolor{b l u e}{\text{This is undefined" => x=4 " is an asymptote}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Point 2}}$

The denominator has ${x}^{3}$ as the most significant figure.
Whilst the numerators most significant figure is $x \sqrt{x}$

Note that $x \sqrt{x} < {x}^{3}$ so as the denominator 'grows' much faster
than the numerator ${\lim}_{x \to \pm \infty} f \left(x\right) \text{ tends to } 0$

$\textcolor{b l u e}{\text{Thus " y=0" is an asymptote}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~