首都師范大學2005年數學分析解（10分求極

2n64.【解答

1 2nlim2 64 lim2 64lim11

n(31164)2 （12分設函

f(x)limx2n1ax2

x2n【解答

1,xx 1ab,x f(xax2bx,1x1f(x在1f(x1a ,x2 ,x 1a 1a故 limaxbx 1， limaxbx x1 a解得b1

x1 （12分已知函數f(x)在(ab上一致連續（ab為有限數。證明：f(x)在(ab上有界【解答f(x)在(ab上一致連續ab為有限數，故對任意0，存在(0,ba)，使得對任xy(ab)xy，就有f(xfy)xy(aa)xy(bbf(xfy)f(af(bf(a),xF(xf(xaxbF(x在[abF(x在(abff(b),x在(ab上有界（10分證明a0n1,2,，且limanl1，則lima0n【解答

naalimanl1，故對任意固定的1,lNN，使得對任意nN，恒有nan，a0，n1,2,，故 an，故對任意nN

0aan1an2aN10，n lima0n（12分f(x在[0,f"(x0f(0)0x0y0，f(xy)f(xfy.【解答函數f(x)在[0,上滿足f"(x)0，故函數f(x)在[0,上連續且可導f(x在[0,f"(x0f'(x在[0,f'(2f'(1f(xy)f(max{x,y})min{x,

f(min{x,y})f(0)min{x,y}x0，y0f(0)0f(xyf(max{xyf(min{xyf(xfy（12分)))M0M2為正常數。證明：對任意x(0,，【解答

f'(x) xh(0,f(xhf(xf'(x)hf"(h2，于是2f'(x)f(xh)f(x)h

f"()h，因此2ff'(x) f"()hf(xh)f(x) h2M0M2ff(xh)f h都成立，2M0M2h的最小值為 x(0,)，有f'(x) （12分f(x在[0,f(x0

2M0M2h ，故對任意 xtf在(0,)內嚴格遞增

(x)

x0f【解答f(x在[0,f(x0x(0, xxx'(x)[0tf(t)dt]'xf(x)0f(t)dtf(x)0tf(t)dtf(x)0(xt)fxxx

0 f(t)dt]' [0f(t)dt]2 [0f(t)dt]2故(x)在(0,內嚴格遞增（12分f(x在[0,1上有二階導1f(0)f(1)0，f"(x)0(x(0,1))，0f(x)dx01證明：(1)f(x在(0,1)(2)存在一點(0,1f'(【解答

f(t)dtf(x)在[0,1上有二階導數，f"(x)0(x(0,1))，故f(x) 0f(x)dx0，故存在c(0,1)f(c0f(x)dx0f(xf(0)f(1)0f(00f(1)0f(x只有一個零點，則在(0,c和1上，f(x)0，故0f(x)dx0 ！故函數f(x)在(0,1)上恰好有兩個零點1xF(x0f(t)dtF"(x)F(x)F"(x)F'(x)F'(x)F(x)[F'(x)F(x)]'[F'(x)F有零點，即證明ex[F'(x)F(x)]'[F'(x)F(x)]{ex[F'(x)F(x)]}'有零點，為此只要證明ex[F'(x)F(x有兩個不同零點，即ex[F'(x)F(x)][exF(x)]'有兩個不同)1G(x)xf(t)dt1f(t)dtxf(t)dt1f(t)dtxf(t)dt2xf

1

0f(t)dtxf(t)dtxf(t)dt2xf充分小鄰域內G(x0，故G(x在(0,1)內有零點F(x在(0,1)內有零點F(x有（10分若級數an與cnanbncn，n證明級數bn【解答anbncn，故0cnbncnanan與cn都收斂，故正項級數(cnan收斂，故正項級數(cnbn)收斂，故級數bn也收斂。（12分I(x

1t2dt(1I(x在(0,limI(x0(2I(x)在(0,)【解答

(1xt(0,，有01t21t20I(x在(0,

1t2dt01t2dt limI(x0

0I(x)

01t

dt

dt 0，xx (2)

x1t2dtt

dta0xat0,01t

0 dt收斂 2在[1,)上單調遞減有界 判別法 2dt在1

1I'(x01t2dt（12分證明函

(xy2)

x2y2f(x,y)

x2 x2y2在(0,0)點可微，但fx(x,y)在(0,0)點不連續f(x,f(x,y)fx2(x,y) 時

x2y2

x2x2x2

(x,y)(0,0)， (x,y

f(xyf(0,0)0f(xy在(0,0)x2fx(xy(x,y)

fy(x,y)0 fx(x,y)2xsin x2 x2 x2(x,y

2x 0，

x2 x2 x2 不存在，

f(xy不存在

(x,y在(x,y)(0,0) x2 x2

(x,y 點不連續（12分Sx2y2z2R2SIS【解答由Gauss公式和輪換對稱性，

5 5I

x3dydz

xdx

x2dxdydz

y2z2)dxdydz x2y2z2 x2y2z2 x2y2z2（12分f(xy在單位圓域x2y21上有連續的偏導數，且在邊界曲線上取值為零。記【解答

I

xfx(x,y)yfy(x,y)x2y2

dxdyxr作廣義極坐標變換