一些簡單高微問題~請求高手或大大幫忙~甘恩~

1.a
lim(x→0)ln(e^x+x)/x
=lim(x→0)[1/(e^x+x)]×(e^x+1)
=1/1×(1+1)
=2
1.b
lim(x→0) (e^x+x)^(1/x)
y=(e^x+x)^(1/x)
lny=(1/x)ln(e^x+x)
y=e^[(1/x)ln(e^x+x)]

lim(x→0) e^[(1/x)ln(e^x+x)]
=e^[lim(x→0)ln(e^x+x)/x]
=e^[lim(x→0)(1/(e^x+x))(e^x+1)]
=e^[1/1(1+1)]
=e^(1×2)
=e^2

2.
f(x)=x^2-1 f(3)=8 6a=8 a=8/6=4/3

3.
i.-2xsinx^2 ii.-2sinxcosx
iii.
(d/dx)∫(x,0) sinte^(-t) dt

∫sinte^(-t) dt
=-∫sintde^(-t)
=-sinte^(-t)+∫e^(-t)dsint
=-sint/e^t+∫cost/e^t dt
=-sint/e^t-∫costde^(-t)
=-sint/e^t-cost/e^t-∫sinx/e^tdt
=[(-1/2e^t)(sint+cost)]∣(x,0)
=(-1/2e^x)(sinx+cosx)-(-1/2e^0)(sin0+cos0)
=(-sinx-cosx/2e^x)+(1/2)(1)
=(-sinx-cosx/2e^x)+(1/2)

(d/dx)∫(x,0) sinte^(-t) dt
=(d/dx)[(-sinx-cosx/2e^x)+(1/2)]
={(-cosx+sinx)2e^x-(-sinx-cosx)(2e^x)}/4e^(2x)
={(-cosx+sinx)2e^x+(-sinx-cosx)(2e^x)}/4e^(2x)
=e^x(-2cosx)/2e^(2x)

4.

x^2-xy+y^2=7
2x-(y+xy')+2yy'=0
2x-y-xy'+2yy'=0
2x-y=xy'-2yy'
2x-y=(x-2y)y'
(2x-y)/(x-2y)=y'

(-2-2)/(-1-4)=4/5


5.
∫(1,-1) (2-∣x∣)dx
=2∫(1,0) (2-x)dx
=2(2x-x^2/2)∣(1,0)
=2(2-1/2)
=4-1
=3

6.

∫e^(-x/2)dx u=(-x/2) du=(-1/2)dx
-2∫e^udu=-2e^u=-2e^(-x/2)
0-(-2)=2

2012-07-27 16:37:49 補充：
不明再問

我已經跟他建議下次這種情形可開三版
每一版只問兩個大題就好

I advise you only ask us to ans one or two Q in each page.

2012-07-27 12:14:32 補充：
Although I know 3 or 4 Q in it, ] I won't ans it.

2012-07-27 19:08:40 補充：
He did 開三版, but he didn't specify us to ans one or two Q.