sin（x）~x
所以 lim(x→0)（1-cos2x)/xsinx
=lim(x→0) (x^2/2)/x^2
=1/2
=lim(x→0) (x^2/2)/x^2
=1/2本回答被提问者采纳
=lim2sinx/x
=2
求极限x趋向于0 (1-cos2x)\/xsinx
用等价无穷小做：当x→0时 1-cos（2x）~(1\/2)x²sin（x）~x 所以 lim(x→0)（1-cos2x)\/xsinx =lim(x→0) (x^2\/2)\/x^2 =1\/2
x趋近于0 求(1-cos2x)\/(xsinx) 的极限
解：lim【x→0】(1-cos2x)\/(xsinx)=lim【x→0】2sin²x\/(xsinx)=lim【x→0】(2sinx)\/x =2 答案：2
lim(x→0)(1-cos2x)\/(xsinx)的极限
求函数lim x→0 1-cos2x\/xsinx 的极限 用等价无穷小代换，x→0时1-cos2x等价于（1\/2)(2x)^2=2x^2,sinx等价于x，lim x→0 1-cos2x\/xsinx =2x^2\/x^2=0
lim(X→o)《1一cos2x》\/xsinx=?
lim(x趋向0)(1-cos2x)\/xsinx =lim(x趋向0)[(1-1+2Sin^2(x)] \/xsinx =lim(x趋向0)2sin^2x\/xsinx =lim(x趋向0)2sinx\/x =2
2.5计算极限lim(x→0) (1-cos2x)\/xsinx
先利用 x等价于sinx有 原式 = lim (1-cos2x)\/x²=lim 2sin2x \/ 2x (洛毕塔法则)=lim 2x\/x =2
limx趋近于0〔1-cos(2x)〕\/(xsinx)
因为cos2x=cos²x-sin²x=1-2sin²x，则有1-cos2x=2sin²x 所以(1-cos2x)\/(xsinx)=(2sin²x)\/(xsinx)=2sinx\/x，根据两个重要极限，得出结果=2
求lim(x→0)(1-con2x)\/xSinx的极限,需要步骤,谢谢!
解：原式=lim(x->0)[2sin²x\/(xsinx)] (应用三角函数倍角公式)=lim(x->0)(2sinx\/x)=2*lim(x->0)(sinx\/x)=2*1 (应用重要极限lim(x->0)(sinx\/x)=1)=2
lim(x→0)(1-cos2x)\/xsinx
lim(x→0)(1-cos2x)\/xsinx=lim(x→0)(1-cosx平方+sinx平方）\/xsinx =lim(x→0)2sinx平方\/xsinx =lim(x→0)2sinx\/x 然后用洛必达 =2lim(x→0) （sinx）'\/x'=2lim(x→0) cosx =2
lim(x→0)(1-cos2x)\/xsinx
记住在x趋于0的时候，1-cosx等价于0.5x^2，而sinx等价于x，那么这里的1-cos2x等价于0.5*(2x)^2 即2x^2 所以得到 原极限=lim(x->0)2x^2 \/ x^2 =2 故极限值为2
求lim(x→∞)1-cos2x\/xsinx
1-cos2x=1-(1-2*(sinX)^2)=2*(sinX)^2,1-cos2x\/xsinx=2sinx\/x 因为当x→∞时,1\/x→0 又sinx为有界函数,|sinx|≤1 所以lim【x→∞】sinx\/x=0 lim(x→∞)1-cos2x\/xsinx=0
