Question

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

Answer 1

`f(x)` 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(x) = (sin x)/(x(x^2-81)) = (sin x)/(x(x-9)(x+9))`

Note that:

• `f(x)` 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->0) (sin x)/(x(x-9)(x+9)) = lim_(x->0) ((sin x)/x * 1/((x-9)(x+9))) = 1 * 1/(-81) = -1/81`
  `""`
  So `f(x)` has a hole at `x=0`.

• When `x=+-9` then numerator is non-zero, so `f(x)` has vertical asymptotes at these values.

• If `x` is real-valued then `abs(sin x) <= 1`. Hence:

  `lim_(x->oo) (sin x)/(x(x^2-81)) = 0`
  `""`
  and `f(x)` has a horizontal asymptote `y=0`

Footnote

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