07级高等数学一上学期期中试题参考解答上学期期中
一求下列极限每小题7分共28分
1lim
→∞






1
2lim
→∞12
解∵lim


解原式lim

→∞



1
11
2∴N0使0
N
→∞12
212
3


lim

→∞
112111

21
∵lim0∴lim0
→∞3
→∞12

3lim
ta
5xx→0l
1xsi
2x
5
4xlimarcta
xsi
→∞
1x
1x
1ta
5x55x解原式lim5x→01si
x10l
1xsi
xxx
解原式limarcta
xlimsi
x→∞x→∞
二求下列函数的导函数每小题7分共28分
1πlimarcta
xsi
lim00x→∞x→∞x2
1yta
x
si
x

2Fx∫x2
x
1tdt求F′x
x
解l
ysi
xl
ta
x
解F′x1
x′
1x2x2′
∴
y′11cosxl
ta
xsi
xyta
xcos2x
si
x

12x
1x2x1x2
∴y′ta
x
1cosxl
ta
xcosx
π3yl
ta
x4
解y′
π′11π′ta
xxπππ44ta
xta
xcos2x4441
1

πxsi
2x2

1
xcos2x



fxtl
tdyd2y求与4tdxdx2ye
dydydxdyet解etl
t1∴dtdtdtdxdxl
t1dt三求下列积分每小题6分共24分dy1etl
t1etdytdx3dxdx2l
t1dt
2
d
1
dt∫si
t
dtdt∫ttt2t2si
cos2ta
cos2222tdta
2l
ta
tC∫t2ta
2
2∫
2x11x2
dx
x1x2dx∫11x2dx
解原式
∫
解原式2
∫
21x2arcsi
xC
23∫l
x1xdx


解原式xl
x1x


2


1∫x
x
1x2dxx1x2xdxxl
x1x21x2C
xl
x1x2∫
1x
2


4
∫x
2
1x
2
dx
22
1xxdx11111解原式∫∫x1xdx∫1xdx∫xdx∫1xdx∫1xdxx1x
22222222
2222
∫
1x
2
1
dx2
xta
t
cos4t12∫cos2tdt∫costdt2∫1cos2tdt

t1112ta
t11xsi
2tCtCarcta
xC224241ta
t221x2
13x∴原式arcta
xCx221x2
四证明不等式
132
≤∫
1
0
x21dx≤31xcosx
7分
f证∵
x2x2x2≤≤21xcosx1
x∈01
∴
132

11112x21xdx≤∫dx≤∫x2dx∫000321xcosx
五设函数gxsi
2xfx其中函数fx在x0处连续
问gx在x0处是否可导若可导求出g′0
解∵
7分
fx在x0处连续∴gxsi
2xfx在x0处有定义且g00
gxg0x0lim
x→0
又∵lim
x→0
si
2xfxr