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

Question

Caleb Alexander · Accepted 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.

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.

Cora Alexander · Answer

undefined To find the vertical, horizontal, and slant asymptotes of the function $ f(x) = \frac{\sin x}{x(x^2 - 81)} $:

1. **Vertical Asymptotes**: These occur where the denominator equals zero, but the numerator does not. Set the denominator equal to zero and solve for $ x $.

2. **Horizontal Asymptotes**: To find horizontal asymptotes, we examine the behavior of the function as $ x $ approaches positive or negative infinity.

3. **Slant Asymptotes**: Slant asymptotes occur when the degree of the numerator is one greater than the degree of the denominator. We perform polynomial long division to check for a slant asymptote. If the result is a non-zero constant, it's a horizontal line which serves as the slant asymptote.

Performing these steps:

1. **Vertical Asymptotes**: The function has vertical asymptotes at $ x = 0 $ and $ x = ±9 $ because these are the points where the denominator becomes zero.

2. **Horizontal Asymptotes**: As $ x $ approaches positive or negative infinity, the function behaves like $ \frac{\sin x}{x^3} $, which approaches zero. Therefore, there is a horizontal asymptote at $ y = 0 $.

3. **Slant Asymptotes**: Perform polynomial long division of $ \sin x $ by $ x(x^2 - 81) $ to see if there's a slant asymptote. The result will give the slant asymptote if it's a non-zero constant. However, in this case, the division yields zero, indicating no slant asymptote.

So, summarizing:
- Vertical asymptotes: $ x = 0, \pm 9 $
- Horizontal asymptote: $ y = 0 $
- No slant asymptote.