lim x→0 (x^4*sin(1/x)+(e^(x^4))-1)/(ln(1+sinx-tanx))
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 22:54:07
lim x→0 (x^4*sin(1/x)+(e^(x^4))-1)/(ln(1+sinx-tanx))
lim x→0 (x^4*sin(1/x)+(e^(x^4))-1)/(ln(1+sinx-tanx))
lim x→0 (x^4*sin(1/x)+(e^(x^4))-1)/(ln(1+sinx-tanx))
lim (x→0)x sin 1/x=
lim(x→0)×(x-sin²3x)/x
lim (x->0) sin(3x)+4x/xsec(x)
求 lim(x→1)sin(x+1)/x+1 求 lim(x→2)x²-4/x-2
求极限 ((sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1
求极限lim(x→0)(1/sin²x-1/x²cos²x)我的算法为什么不对?=lim(x→0)(1/sin²x-1/x²)=lim(x→0)(x²-sin²x / x²sin²x)=lim(x→0)(x²-sin²x /x^4)=lim(x→0)
lim┬(x→0)sinπx/(4(x-1)=
证明lim(x→0)x*sin(1/x)=0我的证明如下:lim(x→0)x*lim(x→0)sin(1/x)因为lim(x→0)x为无穷小量.0
1.lim x→0 3x/(sinx-x)2.lim x→0 (1-cosmx)/x^23.lim x→0 sin4x/[(√x+2)-√2]4.lim h →0 [sin(x+h)-sin(x-h)]/h
lim (x→0+)sin x/x=
LIM(X→∞)X*SIN(1/X)
求极限,lim x趋于0 x * sin 1/x
lim(x趋向0)ln(1+sin x)/x^2
求极限lim(x-->0)x^2 sin(1/x),
lim→0 sin^3/x^3
lim x->0 (sin x-tan x)/sin (x^3)
烦人的数学微积分1、lim(x→0)Sin3x/Sin2x2、lim(x→0)x/3*Sin3x3、lim(x→0)[Xsin(3/x)+(tanx/2x)]4、lim(x→3)Sin(x-3)/(X^2-X-6)5、lim(x→∞)[1+(1/2x)^x6、lim(x→∞)(x+3/x)^x7、lim(x→0)(1-x/2)^1/x+118、求下列函数的间断
求极限 lim sin(x^2 * sin (1/x))/x x->0求极限 lim sin(x^2 * sin (1/x))/x x->0