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

1 Answer
Mar 10, 2016

Answer:

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

Explanation:

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