# How do you find the vertical, horizontal and slant asymptotes of: f(x)=sinx/(x(x^2-81))?

##### 1 Answer
Dec 9, 2017

$f \left(x\right)$ has vertical asymptotes $x = - 9$ and $x = 9$.
It has a hole at $x = 0$ and a horizontal asymptote $y = 0$.

It has no slant asymptotes.

#### Explanation:

Given:

$f \left(x\right) = \frac{\sin x}{x \left({x}^{2} - 81\right)} = \frac{\sin x}{x \left(x - 9\right) \left(x + 9\right)}$

Note that:

• $f \left(x\right)$ is undefined when the denominator is $0$, i.e. when $x = 0$, $x = - 9$ or $x = 9$.

• When $x = 0$ then numerator is also $0$, but we find:

${\lim}_{x \to 0} \frac{\sin x}{x \left(x - 9\right) \left(x + 9\right)} = {\lim}_{x \to 0} \left(\frac{\sin x}{x} \cdot \frac{1}{\left(x - 9\right) \left(x + 9\right)}\right) = 1 \cdot \frac{1}{- 81} = - \frac{1}{81}$
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So $f \left(x\right)$ has a hole at $x = 0$.

• When $x = \pm 9$ then numerator is non-zero, so $f \left(x\right)$ has vertical asymptotes at these values.

• If $x$ is real-valued then $\left\mid \sin x \right\mid \le 1$. Hence:

${\lim}_{x \to \infty} \frac{\sin x}{x \left({x}^{2} - 81\right)} = 0$
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and $f \left(x\right)$ has a horizontal asymptote $y = 0$

Footnote

Historically the line $y = 0$ may not have been considered an asymptote of this $f \left(x\right)$, since the graph of $f \left(x\right)$ crosses it infinitely many times. Modern usage of the term asymptote allows this.