lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/05 17:30:59
![lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2](/uploads/image/z/1907745-33-5.jpg?t=lim%28cosx%2Fcos2x%29%5E%28x%5E1%2F2%29%E6%B1%82%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%2C%E7%AD%94%E6%A1%88%E6%98%AFe%5E3%2F2)
xнN0W*Uj%;6JsNchP12 ++bAsަc<@a9yAI1
du_6e}]!%c?_P071pum/{8Kȇj՞"d*UjbbMp7 453
F\b
lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2
lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2
lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2
lim x→0 (1-cosx√cos2x√cos3x)/(e^x+1)sinx dx
lim[(1-cosx*cos2x****cosnx)/x^2]在x趋于0时
lim cosx(1-√(cos2x))/x^2(当x趋于0)是结果是什么,
高数极限lim[1-cosx(cos2x)^(1/2)]/x^2求解,快
lim(cosx/cos2x)^(x^1/2)求详细过程,答案是e^3/2
求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限
lim (cosx-sinx)/cos2x x→π/4
求极限:lim(x趋向0) [1-cosx(cos2x)^1/2(cos3x)^1/3]/[ln(1+x)-x]
求极限:lim(x→0)(1/x^2)[1-cosx(cos2x)^(1/2)(cos3x)^(1/3)
大一高数 求极限 .lim x->0 [1-cosx(cos2x)^1/2(cos3x)^1/3]/x^2
lim(x趋向0) 1-cos2x/xsinx
lim(x→0)(1-cos2x)/xsinx
lim(cosx+cos2x+ cos3x+…+ cos nx-n)/(cosx-1) (x趋于0)2,3,n之类的是幂,不是2乘以x
(cosx/cos2x)^(1/x^2) 极限
极限x→0 lim[(cosx-cos2x)/x^2]用和差化积的方法
(x趋向于0)lim(cosx-cos2x)/x^2
lim cos2x/(sinx-cosx) x→π/4 求函数的极限
lim(x—0) (1-cosx)/x