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...
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...