How do you find the asymptotes for #f(x)=sinx/(x(x^2-81))#?

1 Answer
Mar 12, 2016

Answer:

Horizontal asymptote:

#y=0#

Vertical asymptotes:

#x=-9#

#x=9#

Explanation:

When we factorize the denominator, we can write

#f(x) = frac{sin(x)}{(x+9)x(x-9)#

For this kind of function, we have to check for the points where the denominator is zero, as there cannot be division by zero. Also we need to check for #+- oo#.

Since the numerator fluctuates about -1 to 1, while the denominator keeps increasing in magnitude, we know that

#lim_{x->oo} f(x) = 0#
#lim_{x->-oo} f(x) = 0#

There is a horizontal asymptote: #y=0#

The denominator equals zero when #x=-9#, #x=0# or #x=9#. We check them one by one.

First, we check the behavior of #f(x)# when #x# is in the region of #-9#. We know that #sin(9) ~~ 0.412 > 0#.

#lim_{x->-9^-} 1/(x(x^2-81)) = oo#
#lim_{x->-9^+} 1/(x(x^2-81)) = -oo#

Therefore,

#lim_{x->-9^-} f(x) = oo#
#lim_{x->-9^+} f(x) = -oo#

There is a vertical asymptote: #x=-9#

Next, we check the behavior of #f(x)# when #x# is in the region of #0#.

Since #f(x)# is of the indeterminate form of #0/0#, we apply the L'hospital Rule.

#lim_{x->0} f(x) = lim_{x->0} frac{sin(x)}{x^3-81x}#

#= lim_{x->0} frac{frac{"d"}{"d"x}(sin(x))}{frac{"d"}{"d"x}(x^3-81x)}#

#= lim_{x->0} frac{cos(x)}{3x^2-81}#

#= frac{cos(0)}{3(0)^2-81}#

#= -1/81#

Seems like #f(x)# is continuous at #x=0# and there is no asymptote.

Now, rather than going through the same process and check for #x=9#, I'm going to say that #f(x)# is an even function. That means

#f(x) = f(-x)#

for every #x# in the domain of #f(x)#. (Try proving this yourself) Therefore, the graph of #y=f(x)# will be symmetrical about the #y#-axis.

Since there is a vertical asymptote of #x=-9#, the other side is going to have its "reflection". The reflection of #x=-9# about #x=0# (the #y#-axis) is #x=9#. Hence, there is a vertical asymptote: #x = 9#

Here is a graph of #y =f(x)# for your reference.
graph{sin(x)/(x^3-81x) [-20, 20, -0.08, 0.08]}