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

##### 1 Answer

#### 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

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:

The denominator equals zero when

First, we check the behavior of

#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:

Next, we check the behavior of

Since

#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

Now, rather than going through the same process and check for

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

for every

Since there is a vertical asymptote of

Here is a graph of

graph{sin(x)/(x^3-81x) [-20, 20, -0.08, 0.08]}