知识大全 lim(x ->-1) ln(2+x)/(1+2x)^1/3+1

Posted 极限

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求:lim(x ->-1) ln(2+x)/(1+2x)^1/3+1

lim(x ->-1) [ln(2+x)]/[(1+2x)^1/3+1]分子分母同时求导
=lim(x ->-1)[(x+2)^(-1)]*[(3/2)*(1+2x)^(2/3)]
=(3/2)lim(x ->-1)[(1+2x)^(2/3)]/(x+2)
x ->-1时分子(1+2x)^(2/3)→1,分母→1
所以=3/2

lim(x→0)(1/ln(1+2x)-1/sin2x)

先通分,然后分母用等价无穷小替换,ln(1+2x)~2x,sin2x~2x,
原式=lim[sin2x-ln(1+2x)]/4x^2=lim[2cos2x-2/(1+2x)]/8x=lim[(1+2x)cos2x-1]/4(1+2x)x
=lim[2cos2x-2(1+2x)sin2x]/4
=2/4
=1/2

求极限ln(1+2x)/sin(1+2x),lim x->0

应该是ln(1+2x)/sin(2x)吧,ln(1+2x)等价于2x,sin(2x)等价于2x,所以极限是1.

lim(x→0)[2^x/㏑(1+2x)-1/㏑(1+2x)]=

lim(x→0) [2^x/ln(1 + 2x) - 1/ln(1 + 2x)]
= lim(x→0) (2^x - 1)/ln(1 + 2x)
= lim(x→0) (2^x - 1)/(2x)、等价无穷小ln(1 + kx) ~ kx
= lim(x→0) [e^(xln2) - 1]/(xln2) * ln(2)/2
= 1 * ln(2)/2
= (1/2)ln(2)
——————————————————————————
其中定理lim(x→0) (e^x - 1)/x = 1
证明:设y = e^x - 1,x = ln(y + 1)
lim(x→0) (e^x - 1)/x
= lim(y→0) y/ln(y + 1)
= lim(y→0) 1/[(1/y)ln(y + 1)]
= lim(y→0) 1/[ln(y + 1)^(1/y)]
= 1/ln(e)
= 1

急求解答! 1、lim x->0 [(1+x)^1/x-(1+2x)^1/2x]/sinx 2、lim x->0 (1/x^2-1/(arctanx)^2) 谢!急!

楼上犯的是典型错误。
1、根据洛必达法则
lim( x->0) [(1+x)^1/x-(1+2x)^1/2x]/sinx =lim (x->0) [(1+x)^1/x]g(x)-[(1+2x)^1/2x]h(x)/cosx
这里g(x)=-(1/x^2)ln(1+x)+1/[x(1+x)]
h(x)=(1/2)[-(1/x^2)ln(1+2x)]+2/[x(1+2x)]
lim( x->0)g(x)= lim( x->0)[-(1/x^2)ln(1+x)]+1/[x(1+x)]=lim( x->0)[-ln(1+x)/2x]=-1/2
lim( x->0)h(x)= lim( x->0)(1/2)[-(1/x^2)ln(1+2x)]+2/[x(1+2x)]
=lim( x->0)(1/2)[-2ln(1+2x)/2x]=-1
∴原式=e(-1/2)-e(-1)=e/2
2、
lim( x->0) (1/x^2-1/(arctanx)^2)
=lim(x->0)(arctanx²-x²)/(x²arctanx²)=lim(x->0)(arctanx²-x²)/(x^4)
=lim(x->0)[(2arctanx)/(1+x^2)-2x]/4x^3=lim(x->0)[(arctanx-x-x^3]/2x^3
=lim(x->0)[1/(1+x^2)]-1-3x^2/6x^2=lim(x->0)(-4x^2-3x^4)/6x^2=-2/3

求lim(x->0)(1+2x)^(1/x)和lim(1-3x)^(1/x)的级限,求步骤!

lim(x->0)(1+2x)^(1/x)
=lim(x->0)(1+2x)^(1/2x)*2
=e^2
lim(1-3x)^(1/x)
=lim(1-3x)^(1/-3x)*(-3)
=e^(-3)
=1/e^3

lim(x→0)(ln(1+2x)/(e^x-1))

0/0型极限,用L\'Hospital法则
lim(x→0)(ln(1+2x)/(e^x-1))
=lim(x→0)((ln(1+2x))\'/(e^x-1)\')
=lim(x→0)(2/(1+2x)/e^x)
=2

x→0 lim(1+2x)^1/x

put y = 1/2x
lim(x->0)(1+2x)^1/x
= lim(y->无限)[(1+1/y)^(y)]^2
= e^2

lim(x→0)=(arctanx-x)/ln(1+2x^3)

说明:此题书写错误,应该是“lim(x→0)[(arctanx-x)/ln(1+2x^3)]”。
解:原式=lim(x->0)[(arctanx-x)\'/(ln(1+2x^3))\'] (0/0型极限,应用罗比达法则)
=lim(x->0)[-(1+2x^3)/(6(1+x^2))]
=-(1+0)/(6(1+0))
=-1/6。

lim√(4x^2+x)ln(2+1/x)-2(lnx)^2=1/4ln2+1

=lim ln((2+1/x)^√(4x∧2+x) /2^x) =lim ln ( (1+1/(2x))^√(4x∧2+x) * 2^(√(4x∧2+x) -2x)) 这里将底数a转换为(a/2)*(2) 再分配出来 =lim ln( ( (1+1/(2x) )^(2x))^(√(4x∧2+x)/(2x)) * 2^(√(4x∧2+x) -2x) ) 构造已知极限e =lim ln (e^(√(4x∧2+x)/。

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