微积分及其应用第二章习题解答

?lim?2??3x4?23x?3x??8

??limx??8?2????x3x2??31?x?3x2??

?4?223x?1?x?32????2.?12.

(4) limx(x?1?x)?limxx???x???2x?1?x2(x?1?x)(5) limx?1?4422x?12x???x?x?x2

2(6) lim(x?x?1?x???x?x?1)

?lim?x22?x?1?x?x?12??2x2? ?1.22x???

?limx???x?x?1?x?x?1?2x?x?1x?x?122(7) limarcsin(x?x?x)?limarcsinx???x???x?x?x2

x?x?x ?limarcsinx???1??arcsin?.

226x?x?xx4 证明方程x?2sinx有非零根.

证明，令f(x)?x?2sinx,易见f(x)在区间?????2?0, ,??上连续，且f()?222???????x0??,??f(?)???0,则由根值存在定理可知存在?2?使得f(x0)?x0?2sinx0?0.即证

方程x?2sinx有非零根

5 证明方程2x3?3x2?e至少有一个正根. 证明 令f(x)?f(1)?5?e?0,2x?3x?e,易见

32f(x)在区间?0,1?上连续，且f(0)??e?0,

x0??0,1?则由根值存在定理可知存在使得

f(x0)?2x0?3x0?e?0.32即

32证方程2x?3x?e至少有一个正根.

复习题二

1已知limf(x)?A，证明lim|f(x)|?|A|.

x?x0x?x0证明：由于limf(x)?A，即对任给的??0,???0,当x?x0??时，有

x?x0f(x)?A??,

x?x0??则对上面给定的??0,???0,当时，有

f(x)?A?f(x)?A??,

lim|f(x)|?|A|即证

x?x0.

， 在极限过程x??下，当a, b为何值时f(x)为无穷小？a, b

2设

f(x)?ax22?2x?1?3bx?5为何值时f(x)为无穷大？ 解 由于

f(x)?ax?2x?122?3bx?5?33bx??5?a?x?3bx?232x?122,

limf(x)?lim3bx??5?a?x?3bx?2x?12要使

x??x???0,

当且仅当b?0,a??5.

3bx??5?a?x?3bx?232limf(x)?lim要使 当且仅当b?0.

x??x??x?12??,

？？？？3? 设x1?0，

limxn?1?a(1?xn)a?xn，a?1，计算n??limxn.

n4设a?b?0，计算n??a?bnn.

nnnn解 由于a?b?0，从而a?a?b?2a,则a??a?bnn?1n1nn??1?2a且lim2na?a,由夹逼

准则可得

limnn??a?b?a.

nn5计算下列极限：

sinxsin（1）lim1xx??x； （2）limxsin12；

x?0x(3)? limx?0?x2sin1cscx2?x??3?e????sinx?2?x???x1?cosxx??tanx?； (4)? lim(2?x)2； ?x?1??5x?13x?12（5）lim(1?x)x?1； （6）lim(x??sin1xx?x?1x2sinx)；

（7）limln(sin22x?e)?x2xxx?0ln(x?e)?2x；

e?e（8）lim；

x?0x?tanxtanx(9)? limxsinln1?x????31； ?sinln1?xx????（10）limx??sinmx. sinnx解 (1) 由于对任意的x都有sinxsin1x?1,即x??时sinxsin1x为有界量，

sinxsinlimx??1x?lim1sinxsin1?0.

x??xxx(2) 由于对任意的x都有sin1x2?1为有界量，故

limxsinx?01x2?0.

(3) 由于对任意的x都有sin1x2?1为有界量,则

xsinlimx?0212xsinx?limxx?0xsinxsin1x2?0.

从而

limx?01?2xsinx2?3?ex?????2?xsinx???????cscx???3?ex???limx?0?2?x???x????cscx1?e?x??xx1?e?x??1?e?x???lim??1???x?0??2?x??????lim2?x2?xcscx

x1?e?x2?x2?x??x?0xx1?e?x?1?e?x?????lim?1??x?0??2?x???????cscx?e.?2?1(4) lim(2?x)x?1tan?2x??1?x?lim?(1?1?x)?x?1??1(1?x)tanx

1lim(1?x)tanx?1?2??1?x ??lim(1?1?x)?x?1??xx2?e?.

2x(5) lim(1?x)1?cosxx?111?cosx?? ?lim?(1?x)x?x?1??1x?11?cosx??2x?e. ??lim(1?x)?x?1??limx2sin1(6) 由于,limx??x?1且当x??,sinx为有界量，则limx?1sinx?0,从而

2x??1xxlim(x??5x?13x?122sin1x?1x?1x2sinx)?lim5x?13x?x2sin

x??x?5.13x太难（7）由于极限是个

00型，由罗必塔法则可得