Question

I am completely lost with this. Any help would be deeply appreciated.

Answer 1

That is the equation. I have worked myself into a circle and I'm seeing double from trying to figure this equation out for the past 20 minutes.

Answer 2

If the discontinuity exists, it's on x = 0, agreed? Otherwise, the function x^2 + x is continuous everywhere.

Answer 3

Do you agree that \(\lim_{x \rightarrow 0^{+}}g(x) = \lim_{x\rightarrow 0^{-}}g(x) = 0\)? Where g(x) = x^2 + x. Now, what does it mean for a discontinuity to be removable? It means that in a given neighborhood (or interval), the function approaches the discontinuity coordinate, because then we can "fix" the hole in the graph with a single point. In the given case, there is no discontinuity because the graph does not have a "hole".

Answer 4

Compare the graph of the f function with: http://www.mathwords.com/r/r_assets/r88.gif Also, you may want to check: http://www.mathwords.com/r/removable_discontinuity.htm

Answer 5

So, if I am reading and understanding this correctly, no discontinuity exists because the function satisfies all 3 conditions of continuity?

Answer 6

Yup, limit of g(x) exists and is equal to f(0), so all three conditions are satisfied.

Answer 7

Thank you so much! =]

Answer 8

No problem :-)