→ ncu1319:SORRY= = 125.232.64.85 02/10 16:35
推 Webb17:感謝這位大大的解答 122.127.58.142 02/10 16:44
推 yuyumagic424:原po好強>/////< 140.112.243.42 02/13 10:52
※ 引述《Webb17 (Webb)》之銘言：
: What is the number of points of discontnuity of the fuction
~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~
題目是問"不連續點"的"個數"
: 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
: 我不曉得為什麼在X=1不連續 請各位高手解答
: 答案是B
由方程式可知在x=2時不連續 所以不連續點的個數為一個
故答案是B
: 1/x^2
: lim ---------
: n→∞ sin^2( 4 )
: ---
: x
: 答案是1/4 我覺得是1/16 也請各位高手解答 謝謝
因為 1/x^2 --> 0 as x--> ∞
sin^2(4/x) --> 0 as x--> ∞
根據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
答案應該是1/16沒錯
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