How can I find the oblique asymptote of $y=\frac{x^3-2x^2+x-1}{x^2+1}$?

Question

Does any one has any idea about how to find the oblique asymptote of this curve?

$$y=\dfrac{x^3-2x^2+x-1}{x^2+1}$$

(original problem image)

Answer 1

$$ \left( x^{3} - 2 x^{2} + x - 1 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x - 2 \right) } + \left( 1 \right) $$ so your fraction is $$ x-2 + \frac{1}{x^2 + 1} $$ and, for large $|x|,$ ......

Answer 2

Polynomial division can be written as $$f(x)=d(x)q(x)+r(x),$$ where $f(x)$ is the polynomial, $d(x)$ is the divisor, $q(x)$ is the quotient, and $r(x)$ is the remainder. Rearranging the equation, we can write $$\frac{f(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)},$$ where $\dfrac{f(x)}{d(x)}$ is now a rational function and $q(x)$ is the asymptote provided that $r(x)\neq0$.

Your polynomial division should result in the equation $$x^3-2x^2+x-1=(x^2+1)(x-2)+1\implies\frac{x^3-2x^2+x-1}{x^2+1}=\underbrace{x-2}_\text{asymptote}+\frac1{x^2+1}.$$

Answer 3

If

$\begin{array}\\ f(x) &= \dfrac{x^3-2x^2+x-1}{x^2+1}\\ \text{then}\\ f(x)-x &= \dfrac{x^3-2x^2+x-1}{x^2+1}-x\\ &= \dfrac{x^3-2x^2+x-x^3-x}{x^2+1}\\ &= \dfrac{-2x^2}{x^2+1}\\ \text{and}\\ f(x)-x+2 &= \dfrac{-2x^2}{x^2+1}+2\\ &= \dfrac{-2x^2+2x^2+2}{x^2+1}\\ &= \dfrac{2}{x^2+1}\\ \end{array} $

so $f(x) =x-2+\dfrac{2}{x^2+1} $.

Answer 4

Note

$$\frac{x^3-2x^2+x-1}{x^2+1}= x-2 +\frac1{x^2+1}$$

which indicates that the oblique asymptote is $x-2$.