1-cos2x的等价无穷小

蓬亲届1391lim(x→0)(1 - cos2x)/(x*sin2x) lim(x→0)(xcot2x) lim(x→无穷)(1+(1/2x))^x -
隗静澜13739779987 ______ lim(x→0)(1-cos2x)/(x*sin2x)=lim(x→0)2sin²x/(x*sin2x) 用等价无穷小=lim(x→0)2x²/(x*2x)=1 lim(x→0)(xcot2x)=lim(x→0)x/tan2x 用等价无穷小=lim(x→0)x/2x=1/2 lim(x→∞)(1+(1/2x))^x=lim(x→∞)(1+(1/2x))^(2x/2) 用重要极限lim(x→∞)(1+1/x)^x=e=e^(1/2)=√e

蓬亲届1391求极限x趋向于0 (1 - cos2x)/xsinx -
隗静澜13739779987 ______ 用等价无穷小做:当x→0时 1-cos(2x)~(1/2)x² sin(x)~x 所以 lim(x→0)(1-cos2x)/xsinx =lim(x→0) (x^2/2)/x^2 =1/2

蓬亲届1391单X→0时,无穷小1 - COS2X与X∧2相比是?A.高阶无穷小 B低阶无穷小C同阶非等阶无穷小D等阶无穷小 -
隗静澜13739779987 ______[答案] 当x趋于0的时候, 1-cos2x是等价于0.5*(2x)^2即2x^2的, 所以与x^2相比应该是同阶非等价无穷小 选择答案C

蓬亲届1391lim(x - >0)=(tanx - sinx)/(1 - cos2x) -
隗静澜13739779987 ______[答案] 利用等价无穷小代换,tanx~x,1-cosx~x²/2,1-cosx~2x².lim(x->0)=(tanx-sinx)/(1-cos2x)=lim(x->0)(tanx-sinx)/(1-cos2x)=lim(x→0)tanx(1-cosx)/(1-cos2x)=lim(x→0)x³/(2x²)=0

蓬亲届1391函数极限问题 limx趋向于0 1 - cos2x/x^2 -
隗静澜13739779987 ______[答案] lim(1-cos2x)/x^2 (0/0型用罗比达法则)(x趋于0) =lim2sin2x/2x (x趋于0) 用等价无穷小 =2

蓬亲届1391求极限lim(x→0) cos(1/x) -
隗静澜13739779987 ______ 由于x趋于0时,1/x趋于无穷,cos(1/x)是跳跃的,极限不存在

蓬亲届1391用洛必塔法则 求极限 lim(x趋于0) (1/sinx^2) - (1/x^2) 求解步骤 -
隗静澜13739779987 ______ ^lim(x→0) (1/sinx^21132)-(1/x^2) (通分)5261 =lim(x→0) (x^2-sinx^2)/(x^2sin^2x) (等价无穷小代4102换1653)回 =lim(x→0) (x^2-sinx^2)/(x^4) (0/0,洛必答达法则) =lim(x→0) (2x-2sinxcosx)/(4x^3) =lim(x→0) (x-1/2sin2x)/(2x^3) (0/0,洛必达法则) =lim(x→0) (1-cos2x)/(6x^2) (等价无穷小代换) =lim(x→0) 1/2(2x)^2/(6x^2) =2/6 =1/3

蓬亲届1391用洛必塔法则 求极限 lim(x趋于0) (1/sinx^2) - (1/x^2) -
隗静澜13739779987 ______[答案] lim(x→0) (1/sinx^2)-(1/x^2) (通分) =lim(x→0) (x^2-sinx^2)/(x^2sin^2x) (等价无穷小代换) =lim(x→0) (x^2-sinx^2)/(x^4) (0/0,洛必达法则) =lim(x→0) (2x-2sinxcosx)/(4x^3) =lim(x→0) (x-1/2sin2x)/(2x^3) (0/0,洛必达法则) =lim(x→0) (1-cos2x)/(6x^2) ...

蓬亲届1391求解一道简单的大一微积分题目,题目在图片中 -
隗静澜13739779987 ______ f′(0) = lim(x→0) [f(x) - f(0)] / (x-0) = lim(x→0) [ (1-cos2x)/x] /x = lim(x→0) (1-cos2x)/x² = lim(x→0) (1 - 1 + 2sin²x) /x² = lim(x→0) 2sin²x/x² (用等价无穷小sinx ~ x) = lim(x→0) 2x²/x² = 2

蓬亲届1391(1 - cosxcos2xcos3x)/(1 - cosx)当x趋近于0时的极限 -
隗静澜13739779987 ______ 由三角积化和差公式 cosxcos2xcos3x =(1/2)(cosx+cos3x)xos3x =(1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x 原极限化为(x->0) (1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx) x->0 1-cosx~(1/2)x^2 上式=(1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4...