lim (x^2sin1/x) /sinx
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 17:33:10
x)UШ3*3ԯT2*lqUb[]}Ϧoy6 Ov2$~ODN}
lV~y[UM_|Km@N ă]
lim (x^2sin1/x) /sinx
lim (x^2sin1/x) /sinx
lim (x^2sin1/x) /sinx
x趋于0时,极限为0lim (x^2sin1/x) /sinx=lim [(sin1/x)/(1/x)]*x/sinx=lim [(sin1/x)/(1/x)]=0
趋于无穷大无极限
lim x→0 sin(x^2sin1/x) / x 等于多少?
lim (x^2sin1/x) /sinx
lim x-->正无穷 (x^2 sin1/x) 怎么算?
正无穷lim(x^2+x)sin1/x
lim(x→+∞)(π/2-arctanx)/(sin1/x)
lim(x→+∞)(π/2-arctanx)/sin1/x,
lim(x(sin)^2)
lim(X->0) sin1/X=?
1、lim(x->无穷大) e^x arctanx2、lim(x->0)sinx√1+sin(1/x)3、lim(x->无穷大)【(√x^2+x+1)-【(√x^2-x+1)】4、lim(x->无穷大)((x+{x+(x)^0.5]^0.5}^0.5)/(2x+1)^0.55、lim(x->0)(sin3x+x^2sin1/x)/((1+cosx)x)6、lim(n->无穷大)(2^n)(si
lim(x→0)(x*sin1/x)/x^2极限为什么不存在?lim(x→0)(x*sin1/x)/x^2= lim(x→0)1/x*sin1/x 为什么这两个无穷小量不可以比较呢
lim(2x^2+1/3x-1)•sin1/x. X趋于无穷大,求极限
求极限lim(x→0)[2sinx+x^2(sin1/x)]/ln(1+x)
lim [2x(sin1/x)-(cos1/x)]/cosx 为什么极限不存在?x→0
lim x趋于0 (sinx+x^2sin1/x)/[(1+cosx)ln(1+x)]
求lim(x^3(sin1/x-1/2*sin2/x)) x趋于无穷
X→0,lim(sin1/x+cos1/x)^x
求极限lim(x→2)(x^2-4)sin1/(x-2),
求极限lim{(x^2/sinx)×sin1÷x}x趋向0