Re: [微積] 三角函數的極限值問題
※ 引述《sugar317 (shadow)》之銘言：
: 這種題目我真的很不懂==
: http://ppt.cc/~PBd
: 該如何求解呢
: 上題答案是1-5sinx
: 下題答案是3sin^2(x)cos(x)+1
: 麻煩了
(1)
cos(x+3h) - cos(x-2h) = -2sin(x + h/2)sin(5h/2)
sin(h)/h -> 1 as h -> 0
-2sin(x + h/2)sin(5h/2) -2sin(x + h/2)sin(5h/2)
----------------------- = ------------------------ * (5/2) -> -5sin(x)
h (5/2)h
所以第一題 = 1 - 5sin(x)
(2)
[sin(x+h)]^3 - [sin(x)]^3
--------------------------- -> 3[sin(x)]^2 cos(x)
h
sin(3h) sin(3h)
------ = 3 * ------- -> 3
h 3h
所以第二題答案 = 3cos(x)[sin(x)]^2 + 1
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