: 1.lim √(x^2+2x)－x = : x→∞ √(x^2+2x) - x = x√[1+(2/x)] - x = x[1+(1/x)-(1/2x^2)...] - x = 1 - 1/2x +... lim √(x^2+2x) - x = 1 x→∞ : 2. sin(3x)*cos(2x) : lim ──────── = : x→∞ tan(6x) mathematica: 不存在 : 3.a line pass through (0,0) is tangent to y=x^3+3x+1 上的(a,b) 求a dy b ---- = 3x^2 + 3, --- = 3a^2 + 3, 又b = a^3 + 3a + 1 dx a a^3 + 3a + 1 = 3a^3 + 3a, 2a^3 = 1, a = 1/2^(1/3) : 4. d x : ─ (∫ √(t^2+1) dt ) = : dx 1 原式 = √(x^2 + 1) : 5. d : ─ (ln(ln(ln(secx)))) = : dx secx * tanx tanx 原式 = -------------------------------- = ------------------------- ln[ln(secx)] * ln(secx) * secx ln[ln(secx)] * ln(secx) : ∞ : 6.lim Σ (2i^2-1)/n^2 = : n→∞ i=0 k 2k(k+1)(2k+1) k 2k^3 + 3k^2 + k - 3k 2k^3 + 3k^2 - 2k Σ (2i^2-1)/n^2 = --------------- - ----- = ---------------------- = ------------------ i=0 6n^2 n^2 3n^2 3n^2 2k^3 + 3k^2 - 2k lim lim ------------------ = ∞ (是这样子算吗...?) n→∞ k→∞ 3n^2 : 2i 2 b : 7.lim sin((1+─)+1)─ 转成 ∫f(x)dx 求a,b,f(x) : n→∞ n n a 题目似乎少打一个Σ... 1 原式 = ∫ 2sin(2+2x) dx, b = 1, a = 0, f(x) = 2sin(2+2x), 此题解有无限多种 0 : 8.∫(e^x)sinx dx = ∫(e^x)sinx dx = (e^x)sinx - ∫(e^x)cosx dx = (e^x)sinx - (e^x)cosx - ∫(e^x)sinx dx e^x 所以说: 原式 = -----(sinx - cosx) 2 : π ∞ : 9.I(k)=∫sin(kx)dx 求Σ I(5^n) = : 0 n=0 -cos(πk) 1 1 cos(π5^n) I(k) = -----------+---, I(5^n) = ----- - ------------ k k 5^n 5^n ∞ 5 5 5 Σ I(5^n) = --- + --- = --- n=0 4 4 2 : 10.y=(x√(x-4))/π 对 X轴旋转的体积 因为当x→∞时, y→∞, 所以说体积∞ : 由scuendless 跟 farrenhi 联手讨论出... : tiesto1114补完第九题+美化 -- ※ 编辑: hsnuyi (118.168.235.20 台湾), 01/15/2020 20:52:19