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

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Answer 1 · 2016-03-10T12:01:12

Asymptotes are $x=4" and "y=0$

Write as: $" "f(x)=(3xsqrt(3x))/((x-4)^3)$

$color(blue)("Point 1")$

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

Thus $lim_(x->4) f(x) =lim_(x->4) (3xsqrt(3x))/((x-4)^3) -> (12sqrt(12))/0$

$color(blue)("This is undefined" => x=4 " is an asymptote")$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Point 2")$

The denominator has $x^3$ as the most significant figure.
Whilst the numerators most significant figure is $xsqrt(x)$

Note that $xsqrt(x) < x^3$ so as the denominator 'grows' much faster
than the numerator $lim_(x->+-oo) f(x)" tends to "0$

$color(blue)("Thus " y=0" is an asymptote")$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B

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