Anonymous
Anonymous asked in 科學數學 · 1 decade ago

微積分(數學)的問題!!請幫幫我!!謝謝!!

"Find a value of c that makes the function

f(x) = { 9x−3 sin 3x/ 5x3 , x 不等於 0

c, x 不等於 0 }

continuous at x = 0. Explain why your value of

c works."

請幫幫我!!

謝謝!!

1 Answer

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  • 1 decade ago
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    我用我的方法算...不知道正不正確

    你就看一下吧 以下算是我把前面的極限符號省略

    且X→0 還有 你的題目應該有誤 C ,X = 0

    否則這提用極限是無解的。

    (1) (9x−3) sin ^3x/ 5x^3

    根據極限定義 LIMf(X)˙g(X) = lim f(x)˙lim g(x)

    所以算式如下

    (9x−3) sin ^3x/ 5x^3

    → 3(3x-1) sin ^3x/ 5x^3

    → 3/5 lim (3x-1) sin ^3x/ x^3 (這邊用極限區分,將常數提出)

    → 3/5lim (3x-1) ˙lim sin ^3x/ x^3

    → 3/5lim (3x-1) ˙ lim sinx / x ˙ lim sinx / x ˙ lim sinx / x

    → 3/5˙-1 ˙1˙1˙1 = - 3/5 (由定義得知 lim sinx / x = 1, x→0)

    所以當x = 0 時 c = - 3/5

    Source(s): 我自己
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