附加數學(極限與導數)一問

lim(x->0) (3x+sin2x)/(7x+2sin4x)

=lim(x->0) [(3x+sin2x)/(8x)]/[(7x+2sin4x)/(8x)]

=lim(x->0) [(1/4)(3/2+sin2x/2x)]/[(1/2)(7/4+2sin4x/4x)]

=[(1/4)(3/2+1)]/[(1/2)(7/4+2)]

=1/3//

佢未必識用 L'hospital rule



2009-09-03 19:27:01 補充：
佢地amath 唔會教ga - -

2009-09-03 20:10:00 補充：
其實好簡單ga 姐 ,

if lim (x->a) f(x)= lim (x->a) g(x) = 0

then lim (x->a) f(x)/g(x) = lim (x->a) f'(x)/g'(x)

2009-09-03 20:13:30 補充：
lim(x->0) (3x+sin2x)/(7x+2sin4x)

你見到分子分母代入個陣都正正等於零 , 錦就符合l'hostipal rule 的假設

我地把 f(x)=3x +sin2x and g(x)=7x+2sin4x 放入去 ,

lim(x->0) f(x)/g(x)

= lim (x->0) f'(x)/g'(x)

= lim (x->0) (3+2cos2x)/(7+8cos4x)

=(3+2cos0)/(7+8cos0) =1/3

2009-09-03 20:14:55 補充：
這都是a-level pure math 計 limit ge 快速方法, 所以你未學好正常,

你錦做 maker 係唔會比分你 ga =[

你用黎研究還可以ge=]

As follows:

圖片參考：http://a.imagehost.org/0120/ScreenHunter_01_Sep_03_19_12.gif


2009-09-03 23:11:17 補充：
唔用L'hospital rule的方法:

http://a.imagehost.org/0727/ScreenHunter_02_Sep_03_23_07.gif