Question

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

Answer 1

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]}