Finding the asymptote of the function $f(x) = \frac{x^3 -x^2 +2}{x-1}$

Question

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?

No, $f(x) = h(x) + \mbox{small}.$ There is also a vertical asymptote at $x=1.$ — Will Jagy, Feb 28 '13 at 04:52
bryan, with your approach, the asymptote of $f(x)=x$ would be $y=-x$. If that seems OK to you, it's time to review the meaning of "asymptote". — Gerry Myerson, Feb 28 '13 at 05:03
If you find the answer to your own question, you should post it as an answer and later accept it. Otherwise, the question will be left unanswered... — tomasz, Mar 08 '13 at 13:46

score 1 · Accepted Answer · answered Mar 08 '13 at 13:48

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 $y=x^2$

Finding the asymptote of the function $f(x) = \frac{x^3 -x^2 +2}{x-1}$

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