StevenXClontz’s Twitter Archive—№ 4,884

@matthematician I think it's something like, for all δ>0, there's e(δ) that limits to 0 as δ does, such that |x-x0|<δ implies |f(x)-f(x0)|<e(δ). To get the traditional proof choose δ=e^(-1)(ε) for ε>0.