计算极限lim(x-0),(1-cos2x)/xsin2x 答案 利用等价无穷小,当x->0时,1-cos2x等价于(2x)^2/2=2x^2而sin2x等价于2x所以,原极限值=2x^2/(x*2x)=1 结果二 题目 求极限:limx→01−cos2xxsinx 答案 原式=limx→02sin2xxsinx=limx→02sinxx=2 结果三 题目 求极限limx→01−cos2xxsin2x ...
计算下列极限.(1)limx→0tan5xx(2)limx→0sin2x3x(3)limx→0xcotx(4)limx→01−cos2xxsinx(5)limn→∞2nsinx2n(6)limx→0(1+2x)1x(7)limx→0(1−3x)1x(8)limx→∞(1+1x)3x(9)limx→∞(1−12x)x(10)limn→∞(2n+32n+1)n+1 答案 (1)5.(2)23.(3)1.(4)2.(5)...
解答一 举报 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 解析看不懂?免费查看同类题视频解析查看解答 相似问题 2.5计算极限lim(x→0) (1-cos2x)/xsinx lim(x→0)(1-cos2x)/xsinx :lim(xsin1/x+1/xsinx)...
有解，cos2x=cox^2x-sin^2x,cox^2x+sin^2x=1,所以分子可以变为：1-cos2x=2sin^2x,和分子上的sinx约掉，所以变为lim 2sinx/x=无穷
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
求lim_(x → 0) (1-cos2x)( xsinx )的值.相关知识点: 试题来源: 解析由题可知,1-cos2x=2sin^2xlim_(x → 0) (1-cos2x)( xsinx )=lim_(x → 0) (2sin^2x)(xsinx)=lim_(x → 0) (2sinx)x=lim_(x → 0) (sinx)x=2
原极限=lim(x->0) 2x^2 / x^2 =2故极限值为2 结果一 题目 lim(x↣0)1-cos2x/xsinx求极限 答案 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 相关推荐 1 lim(x↣0)...
2.5计算极限lim(x→0) (1-cos2x)/xsinx 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析解答一 举报cos2x=1-2sin²x(1-cos2x)/xsinx=[1-((1-2sin²x)]/xsinx=2sin²x/xsinx=2sinx/xlim(x→0) (1-cos2x)/xsinx=lim(x→0) 2sinx/x=2 ...
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 1-cos2x=2sin
lim_(x->0) (1-cos2x)/(x sinx)=lim_(x->0) 2 (sinx)^2/(x sinx)=lim_(x->0) 2sinx/x =2 原