Question

What are the asymptotes and removable discontinuities, if any, of `f(x)=(1-x)/(x^3+2x)`?

Answer 1

Please go through the method of finding the asymptotes and removable discontinuity given below.

Explanation:

Removable discontinuity occurs where there are common factors of numerators and denominators which cancel out.

Let us understand this with an example.

Example `f(x) = (x-2)/(x^2-4)`

`f(x) = (x-2)/((x-2)(x+2)`
`f(x)=cancel(x-2)/((cancel(x-2))(x+2))`

Here `(x-2)` cancels out we get a removable discontinuity at x=2.

To find the Vertical Asymptotes after canceling out the common factor the remaining factors of the denominator are set to zero and solved for `x`.

`(x+2) =0 => x=-2`

The vertical asymptote would be at `x=-2`

The horizontal asymptote can be found by comparing the degree of numerator with that of the denominator.

Say degree of numerator is `m` and degree of denominator is `n`

if `m > n` then no horizontal asymptote
if `m = n` then horizontal asymptote is got by dividing the lead coefficeint of numerator by lead coefficient of denominator.
if `m < n` then y = 0 is the horizontal asymptote.

Now let us see the horizontal asymptotes of our example.

We can see the degree of numerator `(x-2)` is 1
We can see the degree of denominator #(x^2-4) is 2

Degree of denominator is more than degree of numerator therefore the Horizontal asymptote is `y =0`

Now let us come back to our original problem

`f(x)=(1-x)/(x^3+2x)`

Numerator `(1-x)`
Degree of numerator `1`
Denominator `(x^3+2x)`
Degree of denominator `3`
Factors of numerator : `(1-x)`
Factors of denominator: `x(x^2+2)`
No common factors between numerator and denominator therefore no removable discontinuity exist.
Vertical asymptote is found by solving `x(x^2+2) = 0`
`x=0` is the vertical asymptote as `x^2+2=0` cannot be solved.

Degree of denominator is greater than degree of numerator there for `y=0` is the horizontal asymptote.

Final answer : `x=0` vertical asymptote; `y = 0` horizontal asymptote