专升本

lim

x 0

1

2

x

2

lim

x 0

( 1)( 1)

22

x

2

( x 1)

2

lim

1

.......................6分2x 02( x 1)

1

2. 解：

x lim 1 x

x

x

1 lim

1 x

x

x

1

1 lim 1 x x

x

1

....................................................6分e

3．解：

y

1 x

22

xarctanx 2

2

1 x

1 xarctanx

1 x2xy

2

3

.........................................................................6分

4. 解等式两端求微分得

左端 d(e

xy

xy

y

2

lnx) d(e) d(ylnx)

2

2

ed(xy) lnxd(y) yedx xe

xy

xy

y

d(lnx)

2

dx

ydy 2ylnxdy

xy

x

右端 d(sin2x) 2cos2xdx由此得yedx xe

xy

ydy 2ylnxdy

xy

2

xy

2

x

2

dx 2cos2xdx

xe 2xylnx

2xcos2x xye y即y .............................................................................6

xe 2xylnx

xy

2

,

2

xy

整理后得dy

2xcos2x xye

y

dx

5. 解：由换元积分法得

arcsinxdx xarcsinx

x x2

xarcsinx x2 C.......................................6

3