求這題微積分的解法和答案^^

Question

ｌｉｍ (sin2X / x^3 + a + b / x^2)=0
x→0
請問a值和b值答案為??
麻煩告訴我你解法的過程
THX

Answer 1

lim x→0 sin2x/x3 + a + b/x2 = 0lim x→0 (sin2x + ax3 + bx)/x3 = 0∵lim x→0 sin2x + ax3 + bx = 0, lim x→0 x3 = 0∴lim x→0 (sin2x + ax3 + bx)/x3 = lim x→0 (2cos2x + 3ax2 + b)/(2x2)　(羅必達法則)∵lim x→0 2x2 = 0 ∴lim x→0 (2cos2x + 3ax2 + b) = 02cos0 + 0 + b = 02 + b = 0, b = - 2lim x→0 (2cos2x + 3ax2 -2)/(2x2) = lim x→0 (-4sin2x + 6ax)/(4x) = lim x→0 (-8cos2x + 6a)/4　(羅必達法則)(-8cos0 + 6a)/4 = 0-8 + 6a = 0a = 4/3結論 a = 4/3, b = - 2

Answer 2

括號能不能清楚一點? 你寫這樣有很多解讀的方式 > <.

Answer 3

lim sin2X/x^3 =
x->0
lim 2cos2X/3x^2 =
x->0
lim - 4sin2X/6x =
x->0
lim - 8cos2X/6
x->0
=(- 4/3)
(用三次羅必達)
所以
lim (a+b/x^2)=4/3
x->0
又a,b為數值=> 若
lim b/x^2 存在=> b=0 =>則 a=4/3
x->0