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

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 $- \propto$

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 $\propto$

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 $y = 1$