ex2-e2-2cosx求极限lim x4 相关知识点: 试题来源: 解析 解:原极限 =lim_(x→0)e^(-2cosx)⋅(e^x-2+2cosx-1)/(x^4)=lim_(x→0)(x^2-2+2cosx)/(x^4)=lim_(x→0)\frac(2 =1/2lim_(x+0)(1-cosx)/(3x^2)=1/(12) ...
解析 【解析】 ex2=1+x2+12!(x2)2+o[(x2)]2=1+x2+12x4+o(x4) 2-2cosx=2-2 × [1-12!x2+14!x4+o(x4)]=x2-112x e2-2cosx=ex2-112x4+o(x4)=1+[x2-112x4+o(x4 limx→0ex2-e2-2cosxx4=limx-01+x2+12x4+o(x4 ...
ex2=1+x2+ 1 2!(x2)2+o[(x2)]2= 1+x2+ 1 2x4+o(x4); 2−2cosx=2−2×[1− 1 2!x2+ 1 4!x4+o(x4)]= x2− 1 12x4+o(x4), e2−2cosx= e x2− 1 12x4+o(x4)= 1+[x2− 1 12x4+o(x4)]+ 1 2![x2− 1 12x4+o(x4)]2+o(x4)= ...
ex2=1+x2+ 1 2!(x2)2+o[(x2)]2= 1+x2+ 1 2x4+o(x4); 2−2cosx=2−2×[1− 1 2!x2+ 1 4!x4+o(x4)]= x2− 1 12x4+o(x4), e2−2cosx= e x2− 1 12x4+o(x4)= 1+[x2− 1 12x4+o(x4)]+ 1 2![x2− 1 12x4+o(x4)]2+o(x4)= ...
A：当x→0时，（1+sinx）x-1→0．因为（1+sinx）x-1=（1+x+o（x））x-1，令f（x）=（1+x）x，则lnf（x）=xln（1+x），从而，f′(x)f(x)=ln（1+x）+x1+x，f′（x）=f（x）[ln(1+x)+x1+x]=xx[ln(1+x)+x1+x]，f′（0）=0．又因为f″（x）=f′（x）[ln...
计算limx→0ex2−e2−2cosxx4. 答案 ex2=1+x2+12!(x2)2+o[(x2)]2=1+x2+12x4+o(x4);2−2cosx=2−2×[1−12!x2+14!x4+o(x4)]=x2−112x4+o(x4),e2−2cosx=ex2−112x4+o(x4)=1+[x2−112x4+o(x4)]+12![x2−112x4+o(x4)]2+o(x4)=1+x2+512x4+o(x4...
ex2−ln(e+x2)=(1+x2+x42+o(x4))−(x2e−x42e2+o(x4))−1=(1−1e)x2+o(x2), 故ex2−ln(e+x2)的阶数为2. D:因为 ∫1−cosx0arctantdt=∫x22+o(x2)0(t+o(t))dt=x48+o(x4), 所以∫1−cosx0arctantdt的阶数为4. ...
,所以,arcsinx=x−x33+o(x3),从而,x−arcsinx=x33+o(x3),故x−arcsinx的阶数为3.C:因为ln(e+x2)=ln(1+x2e)+1,所以,ex2−ln(e+x2)=(1+x2+x42+o(x4))−(x2e−x42e2+o(x4))−1=(1−1e)x2+o(x2),故ex2−ln(e+x2)的阶数为2.D:因为∫1−cosx0arctan...