直接运用等价无穷小代换,然后代入化简
lim(x趋于0)sin5x\/tan6x求极限
直接运用等价无穷小代换,然后代入化简
三角函数的极限怎么求
lim(x→0)tan5x\/x =5lim(x→0)tan5x\/(5x) =5
求极限lim(x→0) tan5x\/x,用公式tanx=sinx\/cosx来求
第二个方括号: sin(5x)\/(5x)→1 所以 lim(x→0) tan5x\/x=5
limx趋近于0sin5x\/tan7x用洛必达法则求极限
1、本题是无穷小除以无穷小型不定式;2、下面的图片解答,按照楼主的要求,运用罗毕达求导法则计算;3、详细解答过程如下:
limsin5x\/tan4x. x→0求极限!!
解:lim(x→0) sin5x\/tan4x =lim(x→0) sin5xcos4x\/sin4x =lim(x→0) sin5x\/sin4x =lim(x→0) (sin5x\/5x)(4x\/sin4x)(5\/4)=5\/4。
lim(x →0)sin5x\/tan2x怎么求,大神帮忙呀!
lim(x →0)sin5x\/tan2x =lim(x →0)(sin5x\/5x·5x\/2x·2x\/tan2x)=lim(x →0)(sin5x\/5x)·lim(x →0)(5x\/2x)·lim(x →0)(2x\/tan2x)=1·5\/2·1 =5\/2
求极限:sin5x分之tan4x,x趋于0要详细步骤
不知道你的知识范围:用特殊极限 lim(x→0) tan4x\/sin5x=lim(x→0) sin5xcos4x\/sin4x =lim(x→0) sin5x\/sin4x=lim(x→0) (sin5x\/5x)(4x\/sin4x)(4\/5)=4\/5
lim(x →0)sin5x\/tan2x怎么求,
lim(x →0)sin5x\/tan2x =lim(x →0)(sin5x\/5x·5x\/2x·2x\/tan2x)=lim(x →0)(sin5x\/5x)·lim(x →0)(5x\/2x)·lim(x →0)(2x\/tan2x)=1·5\/2·1 =5\/2
lim x→0 sin5x\/tan2x,不用洛必塔法则的求法,麻烦带详细解析。_百度知 ...
x→0 lim sin5x\/tan2x =lim sin(5x)\/(5x) * (5x)\/(2x) * (2x)\/sin(2x) * cos(2x)=lim sin(5x)\/(5x) * lim (5x)\/(2x) * lim (2x)\/sin(2x) * lim cos(2x)因为lim(x→0) sinx\/x=1(重要的极限)=1*(5\/2)*1*1 =5\/2 有不懂欢迎追问 ...