Question

高数极限3道题

1题limx→正无穷（1 - 2/3x）^4x+1
2题limx→正无穷（x/1+x）^-2x+1
3题limx→正无穷（x+1/x-1）^x

要下解题过程，非常感谢~
2题是（x除1+x）^-2x+1

Answer 1

1题.原式=lim(x->+∞){[1-2/(3x)]^(4x+1)}
=lim(x->+∞){[1+(-2)/(3x)]^[(3x)/(-2)]}^[(-2)(4x+1)/(3x)]}
=e^{lim(x->+∞){[(-2)(4x+1)/(3x)]} (利用特殊极限)
=e^（-8/3）
2题.原式=lim(x->+∞){[1+(-1)/(1+x)]^(-2x+1)}
=lim(x->+∞){[1+(-1)/(1+x)]^[(1+x)/(-1)]}^[(2x-1)/(1+x)]
=e^{lim(x->+∞)[(2x-1)/(1+x)]} (利用特殊极限)
=e^2
3题.原式=lim(x->+∞){[(x+1)/(x-1)]^x}
=lim(x->+∞){[1+2/(x-1)]^[(x-1)/2]}^[2x/(x-1)]
=e^{lim(x->+∞)[2x/(x-1)]} (利用特殊极限)
=e^2
注：利用特殊极限lim(x->+∞)[(1+1/x)^x]=1

Answer 2

同学你给的题目叫别人怎么做啊, 第二题是（1/x+x）^(-2x+1)
吧,注意该加括号的地方要加:（x+1/(x-1）)^x,（1 - 2/3x）^(4x+1), 要不然别人怎么做啊

Answer 3

1.
lim（1 - 2/3x）^（4x+1）
=lim(1-2/3x)^[(-3x/2)*(-2/3x)*(4x+1)]
=lim[(1-2/3x)^(-3x/2)]^[-2(4x+1)/3x]

lim(1-2/3x)^(-3x/2)=e，lim[-2(4x+1)/3x]=-8/3
∴
lim[(1-2/3x)^(-3x/2)]^[-2(4x+1)/3x]=e^(-8/3)

2.
lim（x/1+x）^(-2x+1)
=lim[(x+1-1)/(x+1)]^(-2x+1)
=lim[1-1/(x+1)]^[(-(x+1))*(-1/(x+1))*(-2x+1)]
=lim{[1-1/(x+1)]^(-(x+1))}^[(2x-1)/(x+1)]

lim[1-1/(x+1)]^(-(x+1))=e,lim(2x-1)/(x+1)=2
∴
lim{[1-1/(x+1)]^(-(x+1))}^[(2x-1)/(x+1)]=e^2

3.
lim（x+1/x-1）^x
=lim[(x-1+2)/(x-1)]^x
=lim[1+2/(x-1)]^[((x-1)/2)*(2/(x-1))*x]
=lim{[1+2/(x-1)]^((x-1)/2)}^[2x/(x-1)]

lim[1+2/(x-1)]^((x-1)/2)=e,lim2x/(x-1)=2
∴
lim{[1+2/(x-1)]^((x-1)/2)}^[2x/(x-1)]=e^2