關於極限問題

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有公式。
所需公式如下:

A) L' Hospital 's rule 洛必達法則
B) Differentiation of Trigonometric Functions
C) Chain Rule

用法:
A)
當一非常數函數是為不定型 (i) 0/0 或 (ii) ∞/∞ ,則
lim x→x0 [F(x)] / [G(x)] = lim x→x0 [f(x)] / [g(x)]
where f(x), g(x) 為F(x), G(x)之導數。

For indeterminate form of non-constant function: (i) 0/0 or (ii) ∞/∞ , then
lim x→x0 [F(x)] / [G(x)] = lim x→x0 [f(x)] / [g(x)]
where f(x), g(x) is the derivative of F(x), G(x).

B1)
d(sin x)/dx = cos x
B2)
d(cos x)/dx = - sin x

C)
dy/dx = dy/du * du/dx



Now for lim x→0 ( cos 2x - cos 6x ) / x^2

As x→0 ,
cos 2x - cos 6x
= cos 0 - cos 0
= 1-1
= 0

x^2
= 0

分子 ( cos 2x - cos 6x ) 與 分母x^2
同樣趨向零 both tends to zero.
so, it is an indeterminate form. ( 0/0 )

Apply L ' Hospital ' s rule ,

(and by chain rule)
d(cos 2x - cos 6x)/dx = [ (-sin 2x)*2 - (-sin 6x)*6 ]
= -2sin2x + 6sin6x

d (x^2) = 2x

Thus, lim x→0 ( cos 2x - cos 6x ) / x^2 becomes
lim x→0 (-2sin2x + 6sin6x) / 2x

Again, indeterminate form in 0/0, apply L ' Hospital ' s rule,
Thus, lim x→0 (-2sin2x + 6sin6x) / 2x becomes
lim x→0 [ -2 ( 2cos2x) + 6 ( 6cos6x) ] / 2

**********
答案:

lim x→0 ( cos 2x - cos 6x ) / x^2 (0/0)
= lim x→0 (-2sin2x + 6sin6x) / 2x (0/0)
= lim x→0 [ -2 ( 2cos2x) + 6 ( 6cos6x) ] / 2
= (-2* 2 + 6*6 ) / 2
= 16

有
sum to product
cosA-cosB = -2sin[(A+B)/2]sin[(A-B)/2]