=e∧limx→0 (lnsinx-lnx)/x^2(这是0/0型，运用洛必达法则）
=e∧limx→0 (cosx/sinx-1/x)/2x
=e∧limx→0(xcosx-sinx)/(2x^2sinx)
=e∧limx→0(cosx-xsinx-cosx)/(4xsinx+2x^2cosx)
=e∧limx→0-xsinx/(4xsinx+2x^2cosx)
=e∧limx→0-sinx/(4sinx+2xcosx)
=e∧limx→0-cosx/(4cosx+2cosx-2xsinx)
=e∧-1/6
limx→0 (sinx\/x)∧(1\/x∧3)
=limx→0 e∧ln（sinx\/x)^(1\/x^2)=e∧limx→0 (lnsinx-lnx)\/x^2(这是0\/0型，运用洛必达法则）=e∧limx→0 (cosx\/sinx-1\/x)\/2x =e∧limx→0(xcosx-sinx)\/(2x^2sinx)=e∧limx→0(cosx-xsinx-cosx)\/(4xsinx+2x^2cosx)=e∧limx→0-xsinx\/(4xsinx+2x^2cosx)=e∧...
(sinx\/x)^1\/x^3 求当x→0时极限
sin x < x < tan x (0<x<π\/2)以下运用夹逼准则证明右极限等于1 上式各项取倒数，得：1\/tan x < 1\/x < 1\/sin x 各项乘以sin x,得：cos x < (sin x)\/x < 1 当x趋向0式，上面不等式中，cos x趋向1 而最右面也是1，由夹逼准则便有 lim sinx\/x=1(x趋向0(+))因为sinx\/x...
x趋于0时sinx\/x^1\/arcsinx^3的极限
= e^[lim<x→0>ln(sinx\/x)\/arcsin(x^3)]= e^[lim<x→0>(1+sinx\/x-1)\/(x^3)]= e^[lim<x→0>(sinx-x)\/(x^4)]= e^[lim<x→0>(cosx-1)\/(4x^3)]= e^[lim<x→0>(-x^2\/2)\/(4x^3)]= e^[lim<x→0>(-1\/8)(1\/x)]lim<x→0+>e^[lim<x→0+>(-1\/...
当x趋近于0时,(sinX\/X)∧1\/X
lim(x趋于0)(1+3sinx)^[1\/(3sinx)3sinx\/x]=lim(x趋于0)[(1+3sinx)^ 1\/(3sinx)]^(3sinx\/x)显然由重要极限得到x趋于0的时候，(1+3sinx)^ 1\/(3sinx)趋于e，而3sinx\/x趋于3，所以得到极限值趋于e^3
求(sinx\/x)^(1\/x²)在x→0时的极限
简单计算一下即可，答案如图所示
lim(sinx\/x)^x^3,x趋于0时,求极限
L =lim(x->0)(sinx\/x)^x^3 lnL = lim(x->0)ln(sinx\/x) \/ (1\/x^3) (0\/0)= lim(x->0)[(x\/sinx)( xcosx -sinx)\/x^2 ] \/ (-3\/x^4)= -lim(x->0)[(x^3\/sinx)( xcosx -sinx) ] \/3 =0 L = 1
lim(x趋于0)(sinx\/x)∧(1\/(1+cosx)) 极限?谢谢!
如下
求极限x趋于0 时(sinx\/x)^(1\/x^2)
x→0 lim (sinx\/x)^(1\/x^2)=lim e^ln (sinx\/x)^(1\/x^2)=e^lim ln (sinx\/x)^(1\/x^2)=lim (cosx-1)' \/ (3x^2)'=lim -sinx \/ 6x 根据重要的极限：lim sinx\/x=1 =-1\/6 由来 与一切科学的思想方法一样，极限思想也是社会实践的大脑抽象思维的产物。极限的思想可以追溯到...
limx→0(sinx\/ x)^(1\/ x^2)怎么解释
lim(x->0)(sinx\/x)^(1\/x^2)=e^(-1\/6)。解答过程如下：x->0 sinx ~ x-(1\/6)x^3 sinx\/x ~ 1- (1\/6)x^2 令：y = (1\/6)x^2 lim(x->0)(sinx\/x)^(1\/x^2)=lim(x->0)(1 - (1\/6)x^2)^(1\/x^2)=lim(y->0)(1 - y)^[1\/(6y)]=e^(-1\/6)...
求极限x趋于0 时(sinx\/x)^(1\/x^2)
考虑 lim ln (sinx\/x)^(1\/x^2)=lim ln(sinx\/x) \/ x^2 =lim ln(1+sinx\/x - 1) \/ x^2 利用等价无穷小：ln(1+x)~x =lim (sinx\/x - 1) \/ x^2 =lim (sinx-x)\/x^3 该极限为0\/0型,利用L'Hospital法则 =lim (sinx-x)' \/ (x^3)'=lim (cosx-1) \/ (3x^2)该...
