Question

lim(x+5/x+1)^x 大一高数求极限！

lim(x+5/x+1)^x 大一高数求极限！一共两题！谢谢！

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Answer 1

见图

追问

第七题的最后一步怎么来的

不对 是倒数第二步的分母怎么来的

追答

Answer 2

sorry,只有七

Answer 3

Answer 4

运用洛必达法则和等价无穷小代换
（7）原式=lim(x->0+) 2x*[√(1+x^2)-√(1-2x^2)]/x(1-cosx)
=lim(x->0+) 2(1+x^2-1+2x^2)/{(x^2/2)*[√(1+x^2)+√(1-2x^2)]}
=lim(x->0+) 12/[√(1+x^2)+√(1-2x^2)]
=6
（6）原式=lim(x->∞) {[1+4/(x+1)]^[(x+1)/4]}^[4x/(x+1)]
=lim(x->∞) e^[4/(1+1/x)]
=e^4

Answer 5

(6)
(x+5)/(x+1)
=1 + 4/(x+1)
let
4/(x+1)= 1/y
x=4y-1

lim(x->∞) [(x+5)/(x+1)]^x
=lim(x->∞) [1 + 4/(x+1)]^x
=lim(y->∞) (1 + 1/y)^(4y-1)
=lim(y->∞) (1 + 1/y)^(4y)
=e^(4)

(7)
lim(x->0+)∫(0->x^2) [√(1+t) -√(1-2t) ] dt / ∫(0->x) t(1-cost) dt (0/0)
=lim(x->0+) 2x[√(1+x^2) -√(1-2x^2) ] / [x(1-cosx)]
=lim(x->0+) 2[√(1+x^2) -√(1-2x^2) ] /(1-cosx)

x->0
√(1+x^2) ~ 1+(1/2)x^2
√(1-2x^2) ~ 1-x^2
2[√(1+x^2) - √(1-x^2)] ~ 3x^2

cosx ~ 1-(1/2)x^2
1-cosx ~ (1/2)x^2

lim(x->0+)∫(0->x^2) [√(1+t) -√(1-2t) ] dt / ∫(0->x) t(1-cost) dt (0/0)
=lim(x->0+) 2x[√(1+x^2) -√(1-2x^2) ] / [x(1-cosx)]
=lim(x->0+) 2[√(1+x^2) -√(1-2x^2) ] /(1-cosx)
=lim(x->0+) 3x^2 /[ (1/2)x^2 ]
=6