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{(1+1/x)^x^2}/e^x当X趋于无穷大时的极限是多少

发布时间:2019-09-17

lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁺
=lim (x+1)·e^[1/(x-1)]
x→1⁺
=2·e^(+∞)
=+∞
lim [(x²-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)]
x→1⁻
=lim (x+1)·e^[1/(x-1)]
x→1⁻
=2·e⁰
=2
[(x²-1)/(x-1)]·e^[1/(x-1)]在x=1处两侧极限不相等
lim [(x²-1)/(x-1)]·e^[1/(x-1)] 不存在
x→1

回复:

解:
E(Y)=E(e^X)
=∫(0到2)【xe^x/4】dx
=1/4∫(0到2)【xe^x】dx
=1/4∫(0到2)【x】de^x
=1/4xe^x|(0到2)-1/4∫(0到2)【e^x】dx
=e²/2-1/4(e²-e^0)
=e²/4+1/4
D(Y)可能下一周才学呀,让我看看书先

有公式
D(X)=E(X²)-[E(X)]²
于是
D(Y)=E(Y²)-[E(Y)]²=E{(e^x)²}-[E(Y)]²=E(e^2x)-[E(Y)]²
求E(e^2x)
E(e^2x)=∫(0到2)【xe^2x/4】dx

=1/8∫(0到2)【2xe^2x】dx
=1/8∫(0到2)【x】de^2x
=1/8xe^x|(0到2)-1/8∫(0到2)【e^2x】d2x
=e^4/4-1/8(e^4-e^0)
=e^4/8+1/8
于是
D(Y)=E(e^2x)-[E(Y)]²

=e^4/8+1/8-(e²/4+1/4)²
=(e²-1)²/16

回复:

I=∫[-1,1] x⁴/(1+e^x)dx
令t=-x
则 I=∫[1,-1] t⁴/[1+e^(-t)] d(-t)
=∫[-1,1] t⁴/[1+e^(-t)] dt
=∫[-1,1] t⁴ e^t/(1+e^t) dt
=∫[-1,1] x⁴ e^x/(1+e^x)dx
所以
I=1/2∫[-1,1] x⁴/(1+e^x)dx+1/2∫[-1,1] x⁴e^x/(1+e^x)dx
=1/2∫[-1,1] x⁴dx=∫[0,1] x⁴dx=1/5

回复:

解:∫(-π/4到π/4) (cosx)²/[1+e^(-x)]dx =∫(-π/4到0) (cosx)²/[1+e^(-x)]dx+∫(0到π/4) (cosx)²/[1+e^(-x)]dx 对第一个积分式,令t=-x代换下,有: ∫(-π/4到0) (cosx)²/[1+e^(-x)]dx ( t=-x,则dx=-dt) =∫(π/4到0)...

回复:

Ex=∫X1/2*e^(-|x|)dx=0(奇函数,对称区间积分为0) EX²=∫X²1/2*e^(-|x|)dx ,DX=EX²-E²X

回复:

令t=x-1, 则x=t+1, 展开成t的幂级数即可。 e^x=e^(t+1) =e*e^t=e[1+t+t²/2!+t³/3!+......] =e+et+et²/2!+et³/3!+..... 收敛域为R.

回复:

积分法。f(x)=1/(1/2-0)=2 E(2X²)=∫(0,1/2)2x^²f(x)dx =4 ∫(0,1/2)x^²dx =(4/3)[x³] (0,1/2) =(4/3)(1/8) =1/6

回复:

Z=(2X-Y+1)²=4X²-4XY+Y²+4X-2Y+1 EX²=DX+(EX)²=1+1=2 EY²=DY+(EY)²=4+4=8 ρ(XY)=Cov(X,Y)/√(DXDY)=(EXY-EXEY)/2 从而EXY=2ρ(XY)+EXEY=1.2+2=3.2 从而EZ=4EX²-4EXY+EY²+4EX-2EY+1 =8-12.8+8+4-4+1 ...

回复:

x->0时,cosx=1-x²/2!+x^4/24+o(x^4),e^{-x²/2}=1-x²/2+(-x²/2)²/2!+o(x^4)=1-x²/2+x²/8+o(x^4) 所以cosx-e^{-x²/2}=-x^4/12+o(x^4)~-x^4/12 ln(1-x)=-x+x²/2+o(x²),所以x²[x+ln(1-x)]=x...

回复:

lim [(x²-1)/(x-1)]·e^[1/(x-1)] x→1⁺ =lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)] x→1⁺ =lim (x+1)·e^[1/(x-1)] x→1⁺ =2·e^(+∞) =+∞ lim [(x²-1)/(x-1)]·e^[1/(x-1)] x→1⁻ =lim [(x+1)(x-1)/(x-1)]·e^[1/(x-1)] x→...

回复:

I=∫[-1,1] x⁴/(1+e^x)dx 令t=-x 则 I=∫[1,-1] t⁴/[1+e^(-t)] d(-t) =∫[-1,1] t⁴/[1+e^(-t)] dt =∫[-1,1] t⁴ e^t/(1+e^t) dt =∫[-1,1] x⁴ e^x/(1+e^x)dx 所以 I=1/2∫[-1,1] x⁴/(1+e^x)dx+1/2∫[-1,1] x⁴...

回复:

解: 令√(e^x -1)=t,则e^x=t²-1 x=ln(t²-1) ∫√(e^x -1)dt =∫td[ln(t²-1)] =∫2t²/(t²-1)dt =∫(2t²-2+2)/(t²-1)dt =∫2dt +∫[2/(t²-1)]dt =2t+∫[1/(t-1)-1/(t+1)]dt =2t+ln|t-1|-ln|t+1|+C =2t+ln|(t-1)/...

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