Finding the asymptote of a function Find the asymptote of $f(x)$ as $x \rightarrow \infty$ where
$$f(x) = \frac{x^3 -x^2 +2}{x-1}$$
The answer key mentioned that the answer is $h(x)=x^2$ where $h(x)$ is the equation of the asymptote of $f(x)$. I beg to differ with this solution:
$$
\begin{align*}
f(x) = \frac{x^3 -x^2 +2}{x-1} 
&= \frac{x^2 (x-1) +2}{x-1}\\
&=x^2+\frac{2}{x-1}
\end{align*}
$$
With arrangement, we attempt to fit it in the form $f(x) + h(x) =\frac{2}{x-1}$ and so we have
$h(x) = -x^2$. So the asymptote of $f(x)$ as $x \rightarrow \infty$ is $y=-x^2$
Any ideas to solve this?

With a rearrangement, 
$$
\begin{align*}
f(x) -x^2
&=\frac{2}{x-1}\\
f(x) -h(x)
&=\frac{2}{x-1}\\
\lim_{x\rightarrow\infty}f(x) -h(x)
&=\lim_{x\rightarrow\infty}\frac{2}{x-1}\\
&=0
\end{align*}
$$

If you think about it, the idea is that as $x \rightarrow \infty$, the difference between $f(x)$ and $h(x)$ tends to $0$. So the functions are getting closer and closer to each other, but don't touch. It is no wonder $h(x)$ is considered the asymptote.

Hence, indeed the asymptote is $y=x^2$
