※ 引述《Webb17 (Webb)》之铭言： : What is the number of points of discontnuity of the fuction ~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~ 题目是问"不连续点"的"个数" : 1 : x+ ─ ,if x<=0 : 2 : f(x)= 1 : ─ ,if 0<x<=2 : 2 : 3-x ,if 2<x<=3 : (3-x)^2 ,if 3<x : (a)0 (b)1 (c)3 (d)4 : 我不晓得为什麽在X=1不连续 请各位高手解答 : 答案是B 由方程式可知在x=2时不连续 所以不连续点的个数为一个 故答案是B : 1/x^2 : lim --------- : n→∞ sin^2( 4 ) : --- : x : 答案是1/4 我觉得是1/16 也请各位高手解答 谢谢 因为 1/x^2 --> 0 as x--> ∞ sin^2(4/x) --> 0 as x--> ∞ 根据L'Hopital's rule 1/x^2 (-2)/x^3 lim ------------- = lim ------------------------------ x-->∞ sin^2(4/x) x-->∞ 2sin(4/x)*cos(4/x)*4*(-1/x^2) 1/x = lim ---------------------- x-->∞ 4*sin(4/x)*cos(4/x) (-1)/x^2 = lim ----------------------------------------------- x-->∞ 4[cos(4/x)]^2*4*(-1/x^2)-4[sin(4/x)]^2*4*(-1/x^2) 1 = lim ------------------------------------- x-->∞ 16 [cos(4/x)]^2 - 16[sin(4/x)]^2 = 1/16 答案应该是1/16没错 --

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