limit一條數唔識..有冇人可以幫~

lim (sin3x-3sinx)/x^3
x->0
sin3x
= sin(2x + x)
= sin2xcosx + sinxcos2x
= 2sinxcos^2(x) + sinx[1 - 2sin^2(x)]
= 2sinx - 2sin^3(x) + sinx - 2sin^3(x)
= 3sinx - 4sin^3(x)
SO,
lim (sin3x-3sinx)/x^3
x->0

= lim [ - 4sin^3(x)]/ x^3
x->0

= -4<------for lim[x->0] {sin^3(x) / x^3}= 0




2007-04-03 17:50:48 補充：
for the prove of sin3xi use the trigonometry formulasin(A 十 B) = sinAcosB 十 sinBcosA

firstly sin 3x = 3sin(x) - 4 sin^3(x)
there fore :
lim 3sin(x) - 4sin^3(x) - 3sin(x) /x^3
= -4* (sin^3(x) / x^3)
as x tends to 0 sinx tends to x
so
-4* (x^3/x^3)
= -4*1
= -4