How do you find vertical, horizontal and oblique asymptotes for f(x) =(x-1)/(x-x^3)?

Jul 21, 2017

The line $x = 0$ is a vertical asymptode of $f$

The line $x = - 1$ is a vertical asymptode of $f$

So the line $y = 0$ is the horizontal asymptode of $f$ in both $+ \infty$ and $- \infty$

Explanation:

$f \left(x\right) = \frac{x - 1}{x - {x}^{3}} = \frac{x - 1}{x \left(1 - {x}^{2}\right)} = \frac{x - 1}{x \left(1 - x\right) \left(1 + x\right)}$

Now let's find first the vertical asymtodes of $f$ :

${\lim}_{x \rightarrow 1} f \left(x\right) = {\lim}_{x \rightarrow 1} \frac{x - 1}{x \left(1 - x\right) \left(1 + x\right)} =$

${\lim}_{x \rightarrow 1} - \frac{\cancel{1 - x}}{x \cancel{\left(1 - x\right)} \left(1 + x\right)} = {\lim}_{x \rightarrow 1} - \frac{1}{x \left(1 + x\right)}$

$- \frac{1}{1 \cdot 2} = - \frac{1}{2}$

Because $- \frac{1}{2} \in \mathbb{R}$ there is no vertical asymtode at $x = 1$

${\lim}_{x \rightarrow {0}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {0}^{+}} - \frac{1}{x \left(1 + x\right)} = - \infty$

So the line $x = 0$ is a vertical asymptode of $f$

${\lim}_{x \rightarrow - {1}^{+}} f \left(x\right) = {\lim}_{x \rightarrow - {1}^{+}} - \frac{1}{x \left(1 + x\right)} = - \infty$

Also the line $x = - 1$ is a vertical asymptode of $f$

Now for the oblique and horizontal ones :

If ${\lim}_{x \rightarrow \pm \infty} f \frac{x}{x} = m \in \mathbb{R}$ and ${\lim}_{x \rightarrow \pm \infty} \left(f \left(x\right) - m x\right) = b \in \mathbb{R}$

Then the line $y = m x + b$ is an oblique asyptode at $\pm \infty$ respectivly

If $m = 0$ then the asymptode is horizontal

${\lim}_{x \rightarrow \pm \infty} f \frac{x}{x} = {\lim}_{x \rightarrow \pm \infty} \frac{x - 1}{{x}^{2} - {x}^{4}} = {\lim}_{x \rightarrow \pm \infty} \frac{x}{- {x}^{4}} = 0$

${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = {\lim}_{x \rightarrow \pm \infty} \frac{x - 1}{x - {x}^{3}} = 0$

So the line $y = 0$ is the horizontal asymptode of $f$ in both $+ \infty$ and $- \infty$

Jul 21, 2017

$\text{vertical asymptote at " x=0" and } x = - 1$
$\text{horizontal asymptote at } y = 0$

Explanation:

$\text{simplifying f(x)}$

$f \left(x\right) = \frac{x - 1}{x \left(1 - {x}^{2}\right)} = - \frac{\cancel{\left(1 - x\right)}}{x \cancel{\left(1 - x\right)} \left(1 + x\right)}$

$\Rightarrow f \left(x\right) = - \frac{1}{x \left(1 + x\right)}$

$\text{the removal of the factor "(1-x)" from the }$
$\text{numerator/denominator indicates a hole at } x = 1$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve "x(1+x)=0rArrx=0" and } x = - 1$

$\Rightarrow x = 0 \text{ and " x=-1" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

divide terms on numerator/denominator by the highest power of x. that is ${x}^{2}$

$f \left(x\right) = - \frac{\frac{1}{x} ^ 2}{\frac{x}{x} ^ 2 + {x}^{2} / {x}^{2}} = - \frac{\frac{1}{x} ^ 2}{\frac{1}{x} + 1}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{0 + 1}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{-(1)/(x(x+1) [-10, 10, -5, 5]}