证明：函数y = f(x) = x^1/3 在区间(-∞,+∞)内连续,但在点x = 0处不可导.因为在点x = 0处有[f(0+h)-f(0)]/h = (h^(1/3) - 0)/h = 1/h^(2/3)因此极限 lim(h→0) [f(h+0)-f(0)]/h = lim(h→0) 1/h^(2/3) = +∞即导数...