For absolute value of x, f(x) = lxl To prove continuous, I can express it as a piecewise function of x if x>0 and -x if x<0 and then I can take the limit of each as x approaches 0 from the negative and positive sides. Both limits give zero, and f(0) gives zero, so continuity is proven, right? Also, if I take the limit as h approaches 0 and get the general derivative function of x/lxl , whatever number I plug into that derivative function is going to be the slope of the tangent at a specific point. So if I pick x = 5 so that f'(5) = 1, so the slope is 1, and if I pick -5, the slope will be -1. However, I'm confused at x = 0. for the original function lxl, there is no denominator so no vertical asymptote. But for the derivative function x/lxl there is a denominator, and the denominator can't be zero. That means there must be a vertical asymptote at x = 0 on the function x/lxl. Now, if I plug 0 into the derivative, f'(0) = 0/l0l, I get 0/0 which is no solution. So which condition, specifically, determines that the function is not differentiable at x = 0? The fact that the denominator of f prime can't be zero, or that f prime of zero produces no solution?