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

1 Answer
Dec 9, 2017

Answer:

#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.