How do I find the vertical and horizontal asymptotes of #f(x)=x^2/(x-2)^2#?

1 Answer
Jun 1, 2015

These are the illegal values for x
You mustn't divide by 0, thus #(x-2)²# mustn't be equal to 0
Which means #x-2# mustn't be equal to 0
And thus #x# mustn't be equal to 2
2 is the illegal value

So there is a vertical asymptote in #x=2#

  • Horizontal asymptotes :

We are looking for the limits

  1. Limit in #-prop#

#lim(x²)=+prop#
#lim(x-2)²=lim(x²-4)=+prop#
The #-4# is insignificant since we deal with really great numbers
So #limf(x)=lim((x²)/(x²))=lim(1)=1#

We thus have an horizontal asymptote in #y=1#

  1. Limit in #prop#

#lim(x²)=+prop#
#lim(x-2)²=lim(x²-4)=+prop#
The #-4# is insignificant since we deak with really great numbers
So #limf(x)=lim((x²)/(x²))=lim(1)=1#

Same vertical asymptote in #y=1#