sinx-tanx的极限怎么求

毕郊咐4945(Sinx - tanx)/4xx+tanx趋近0的极限 -
匡览研13158635584 ______[答案] 题目若是 lim[(sinx-tanx)/(4x^2 )+tanx ] =lim[tanx(cosx-1)/(4x^2 )+tanx ] =lim[(-x^2/2)*tanx/(4x^2 )+tanx ] =lim[-tanx/8+tanx ]=0 题目若是 lim[(sinx-tanx)/(4x^2 +tanx)] =lim[tanx(cosx-1)/(4x^2 +tanx)] =lim[(-x^3/2)/(4x^2 +tanx)] =0 题目 若是 lim[(sinx-tanx)/(4x^2 *...

毕郊咐4945当x趋向于0时,(sin(x) - tan(x))/x^3 的极限怎么求? -
匡览研13158635584 ______ 解:∵(sinx-tanx)/x³=(sinx/x)*[(cosx-1)/x²]*(1/cosx) 又lim(x->0)(sinx/x)=1 (这是重要极限,要记熟) lim(x->0)[(cosx-1)/x²]=lim(x->0)[(-sinx)/(2x)] (0/0型极限,应用罗比达法则) =-1/2lim(x->0)(sinx/x) =-1/2 (应用重要极限) lim(x->0)(1/cosx)=1 ∴原式=lim(x->0)(sinx/x)*lim(x->0)[(cosx-1)/x²]*lim(x->0)(1/cosx) =1*(-1/2)*1 =-1/2.

毕郊咐4945求极限(x→0)(sinx - tanx)/(x^2(e^2x - 1)) -
匡览研13158635584 ______ 你好!先用等价无穷小:e^(2x) -1 ~ 2x 原式 =lim<x→0> (sinx - tanx) / (2x^3)=lim<x→0> [cosx - (secx)^2] / 6x^2 【罗比达法则】=lim<x→0> [(cosx)^3 -1] / [6x^2 (cosx)^2)]=lim<x→0> (cosx -1)[(cosx)^2 +cosx +1] / [6x^2 (cosx)^2) ]=lim<x→0> [(cosx)^2 +cosx +1] / [12 (cosx)^2 ] 【等价无穷小cosx -1 ~ 1/2 x^2】= 1/4

毕郊咐4945当x趋向于0时,(sinx - tanx)/arcsin^3x的极限 -
匡览研13158635584 ______ arcsin^3x与x^3是等价无穷小 lim(sinx-tanx)/arcsin^3x =lim(sinx-tanx)/x^3 (下面用罗比达法则) =lim(cosx-(secx)^2)/3x^2 =lim((cosx)^3-1)/(3x^2(cosx)^2) =lim((cosx)^3-1)/(3x^2) =lim3((cosx)^2*(-sinx))/(6x) =-1/2

毕郊咐4945原题是lim x趋向于0 sinx - tanx除以4x^2+tanx 求极限 原题是lim x趋向于0 sinx - tanx除以4x^2+tanx 求极限 -
匡览研13158635584 ______[答案] lim (x->0)( sinx-tanx)/(4x^2+tanx) =lim (x->0)( x-x)/(4x^2+x) =0

毕郊咐4945x→0时,sinx - tanx/{[(1+x^2)^1/3 - 1][(1+sinx)^1/2 - 1]的极限为什么等于 - 3 -
匡览研13158635584 ______[答案] 首先,由等价无穷小替换 分母=1/3x^2*1/2x=1/6x^3 所以原式=(极限符号略去) (sinx-tanx)/(1/6x^3) =6sinx(1-secx)/x^3 =6(1-secx)/x^2 =6(-secxtanx)/(2x) =-3(secx) =-3

毕郊咐4945(sinx - tanx)/x^3 x趋于0 的极限怎么求?答案是 - 0.5 可不可以给出过程?麻烦赐教了! -
匡览研13158635584 ______[答案] 答案是 -0.5 (sinx-tanx)/x^3 = [sinx(1-1/cosx)]/x^3= [(sinx)/x]*(1-1/cosx)/x^2当x趋近于0时,(sinx)/x=1所以lim(sinx-tanx)/x^3 = lim(1-1/cosx)/x^2当x趋近于0时,(1-1/cosx)/x^2 的分子分母都趋近于0,所以用罗比...

毕郊咐4945当X趋向于0时求(sinx - tanx)/[(根号下1+X^2) - 1][(根号下1+sinX) - 1]的极限 -
匡览研13158635584 ______[答案] Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1) (Sqrt[1 + Sin[x]^2] - 1), x -> 0] 等量无穷小. =Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1)^2, x -> 0] 应用洛必达法则 = Limit [(Cos[x] - Sec[x]^2)/((2 x (-1 + Sqrt[1 + x^2]))/Sqrt[1 + x^2]), x -> 0] 化简 = Limit [((Cos[x]^3 - 1)/...

毕郊咐4945求x趋向于π/2时,(sinx)^tanx的极限 -
匡览研13158635584 ______ 解:当u->0时 ,(1+u)^(1/u) -> e 当x->π/2 时,令 u = sinx-1,u->0 (sinx) ^ (tanx) = (1+ sinx-1) ^ (tanx) = (1+u) ^ {(1/u) * u * tanx } lim(x->π/2) u * tanx 令 t = π/2 -x = lim(t->0) (cost - 1)/ tant = lim(t->0) (cost - 1)/ t = 0 故 lim(x->π/2) (sinx) ^ (tanx) = e^0 = 1 ...

毕郊咐4945lim x趋向派 sinx/tanx的极限,求解 -
匡览研13158635584 ______ sinx/tanx=sin(π-x)/tan(π-x)~(π-x)/(π-x) 所以lim( x→π )sinx/tanx=1