Local maxima/minima and removable discontinuities

f'(x) = x^2 = 0 implies a global minimum at x = 0, but you cannot forget that the meaning of derivatives are true only for delta x --> 0. Just because a global minimum exists by analysis of derivatives does not mean that it actually exists! What about x^2/x? That's even more problematic.

214 got it right, or nearly - you have a "local minima" if you pick a point very very close to (but not equal to) 0, but it is not a true minima because I can argue by a next-step-closer-towards-0 PoV, which just goes back to the meaning of "infinity". By l'Hopital's rule the limit clearly exists at x = 0, but if the function fails to exist at x = 0, then there clearly isn't a local minima.

My 2 pennies. Generally I shy away from math where possible, especially analysis.

Hadriel