# What are the asymptote(s) and hole(s), if any, of # f(x) =((x-3)(x+2)*x)/((x^2-x)(x^3-3x^2) #?

##### 1 Answer

#### Explanation:

First, let's simplify our fraction without cancelling anything out (since we're going to be taking limits and cancelling stuff out might mess with that).

#f(x) = ((x-3)(x+2)(x))/((x^2-x)(x^3-3x^2))#

#f(x) = ((x-3)(x+2)(x))/((x)(x-1)(x^2)(x-3))#

#f(x) = (x(x-3)(x+2))/(x^3(x-1)(x-3)#

Now: holes and asymptotes are values which make a function undefined. Since we have a rational function, it will be undefined if and only if the denominator is equal to 0. We therefore only need to check the values of

#x=0#

#x=1#

#x=3#

To find out whether these are asymptotes or holes, let's take the limit of

#lim_(x->0)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = lim_(x->0)((x-3)(x+2))/(x^2(x-1)(x-3))#

#= (-3*2)/(0*(-1)*(-3)) = +-oo#

So

#lim_(x->1)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = (1*(-2)*3)/(1*0*(-2)) = +-oo#

So

#lim_(x->3)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = lim_(x->3)((x+2))/(x^2(x-1))#

#= 5/(9*2) = 5/18#

So

*Final Answer*