Recent content by GangSines

I would just have to choose that number such that it makes g(x0) continuous?

My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part

After using L'hospital's rule, you find that the limit is f'(x0) ≠ g(x0). So g is not continuous at x0.

Homework Statement A function f has a simple zero (or zero of multiplicity 1) at x0 if f is differentiable in a neighborhood of x0 and f(x0) = 0 while f(x0) ≠ 0. Prove that f has a simple zero at x0 iff f(x) = g(x)(x - x0), where g is continuous at x0 and differentiable in a deleted...