1. ## Finding Removable Discontinuities

I have the function (x^2+3x)/(x^3-x), and I have to find it's removable discontinuity. I factored it, and got x(x+3)/x(x+1)(x-1). The single x in the top and the bottom cancels out, and solving for zero obviously gives me x=0 for the removable discontinuity. However, whenever I graph the function, I consistently find that the graph is undefined at 0,-3...

2. ## Re: Finding Removable Discontinuities

Originally Posted by qubez
I have the function (x^2+3x)/(x^3-x), and I have to find it's removable discontinuity. I factored it, and got x(x+3)/x(x+1)(x-1). The single x in the top and the bottom cancels out, and solving for zero obviously gives me x=0 for the removable discontinuity. However, whenever I graph the function, I consistently find that the graph is undefined at 0,-3...
You must be entering incorrectly.

3. ## Re: Finding Removable Discontinuities

Is there just no discontinuity? There isn't one at x=0, is there?

4. ## Re: Finding Removable Discontinuities

Originally Posted by qubez
Is there just no discontinuity? There isn't one at x=0, is there?
Well there are three discontinuities: $\displaystyle -1,~0,~1$.
But $\displaystyle 0$ is a removable discontinuity; $\displaystyle \pm 1$ are essential discontinuities.

5. ## Re: Finding Removable Discontinuities

I guess I'm just confused because there are asymptotes at x=-1 and x=1, so the graph never gets near x=0, so how could you insert a point at x=0 and make the graph continuous? That question probably made no sense...

6. ## Re: Finding Removable Discontinuities

Originally Posted by qubez
I guess I'm just confused because there are asymptotes at x=-1 and x=1, so the graph never gets near x=0, so how could you insert a point at x=0 and make the graph continuous? That question probably made no sense...
Are we looking at the same graph? Redefine the function (let's call it $\displaystyle f$) so that $\displaystyle f(0)=-3$. Then it's continuous at $\displaystyle x=0$.

7. ## Re: Finding Removable Discontinuities

Originally Posted by Reckoner
Are we looking at the same graph? Redefine the function (let's call it $\displaystyle f$) so that $\displaystyle f(0)=-3$. Then it's continuous at $\displaystyle x=0$.
Yes $\displaystyle x=0$ is removable as I said. SO what is new?