How do you find the vertical, horizontal and slant asymptotes of: f(x)=sinx/(x(x^2-81))f(x)=sinxx(x281)?

1 Answer
Dec 9, 2017

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

It has no slant asymptotes.

Explanation:

Given:

f(x) = (sin x)/(x(x^2-81)) = (sin x)/(x(x-9)(x+9))f(x)=sinxx(x281)=sinxx(x9)(x+9)

Note that:

  • f(x)f(x) is undefined when the denominator is 00, i.e. when x = 0x=0, x = -9x=9 or x = 9x=9.

  • When x=0x=0 then numerator is also 00, 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.