Find asymptotes of $(2x)/(x-1)^2$ What are the asymptotes of $$\frac{2x}{(x-1)^2}$$ ? 
I have problems already on domain.
 A: Asymptote: $\displaystyle{y = ax + b}$.
$$
\lim_{x \to \infty}\left[{2x \over \left(x - 1\right)^{2}} - ax - b\right] = 0\,,
\quad
\lim_{x \to \infty}
\left\{x\left[{2 \over \left(x - 1\right)^{2}} - a - {b \over x}\right]\right\}
=0
$$
$$
\lim_{x \to \infty}
\left[{2 \over \left(x - 1\right)^{2}} - a - {b \over x}\right] = 0
\quad\Longrightarrow\quad
a = 0
$$
$$
\lim_{x \to \infty}
\left[{2x \over \left(x - 1\right)^{2}} - b\right] = 0
\quad\Longrightarrow\quad
b = 0
$$
$$
\mbox{Asymptote:}\quad y = 0
$$ 
A: So $f(x)=\frac{2x}{(x-1)^2}$. Here you horizontal asymptote will be found by taking the lim as x $\rightarrow\infty$, which should be $0$. And your veritval asymptotes will be where the denominator vanishes (i.e when $ (x-1)^2=0$).
A: There are three types of asymptotes you will need to check for: 


*

*$\textbf{Vertical asymptotes}$: These are the $x$-values at which the denominator evaluates to zero. In your case, the vertical asymptote is at $x=1$.

*$\textbf{Horizontal asymptotes}$: If you rational function is proper, i.e the degree of the numerator is lower than the degree of the denominator, then the horizontal asymptote will be at $y=0$. Notice,that your rational function is $\textbf{not}$ proper. 
To find the horizontal asymptote of a rational function which is not proper, as is the case here, you need to use polynomial long division, i.e divide $(x-1)^2$ by $x$. The quotient is a horizontal asymptote if it is a constant.


*$\textbf{Oblique Asymptote}$: If your quotient is a linear function, i.e., of the form $ax+b$, then you have an oblique asymptote. 

