四、计算题
2n2?3n?21.lim. 2n??2n?132122??2(2n?3n?2)22n?3n?2nn 解 lim=limn=lim2n??n??n??112n?12(2n?1)2?n2n2322?lim?lim22?0?0n??nn??n?1 ==
12?02?lim2n??n22.求极限lim(n1?n2???n10).
n??解 利用数列极限四则运算公式有
lim(n1?n2???n10)?limn1?limn2???limn10=10
n??n??n??n??3.求极限lim4?8x????xx?1x.
1xx解 limx????x4?x2??1??1?8lim8???1lim8??1?x???x????xx???2??2????1x????11?2xx?8?e0?8
4.y?ln1x?x2?1，求y'.
解 首先y??1lnx?x2?1，再求导有 2111y??(?lnx?x2?1)?????(x?x2?1)?
22x?x2?1112x1 ????(1?)??2222x?x?12x?12x?15. 设y?2ln1?x?1?x 求y'.
1?x?1?x解 首先将函数变形
?y?2ln根据函数求导法则
1?x?1?x2x?2?4ln?1?x?1?x?2lnx?2ln2，
?y'?2?1?x??12??1?x??121?x?1?x?22. ?2xx1?x?5x2,6.求函数f(x)???ax?b,
x?c,x?c.在c的右导数. 当a与b为何值时, 函数f(x)在c可导.
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5x2?5c2?10c.若可导, 则f(x)在c点连续, 所以f(c?0)?f(c?0). 于是有5c2?ac?b.解 f??(c)?lim?x?cx?c并且f??(c)?f??(c), 所以f??(c)?lim?ax?b?(ac?b)?a?10c. 所以a?10c,b??5c2.时, 函数f(x)在c可导.
x?cx?c1111.求极限：lim(1?2)(1?2)?(1?2).
n??23n解 注意1?1(n?1)(n?1)?，则 22nn1111*32*4(n?1)(n?1)limn??(1?22)(1?32)?(1?n2)=limn??(22*32? =12232?(n?1)2n(n?1)2limn(n?1)n??2232?(n?1)2n2=lim1n??2n2=2 2.求极限lim1?2x?3x?4x?2.
解 lim1?2x?3x?4x?2?lim(1?2x?9)(x?2) x?4(x?4)(1?2x?3) ?lim(2x?8)(x?2)2(x?x?4(x?4)(1?2x?3)?lim2)x?4(1?2x?3)?43
3.设y?arctanx2?1?lnxx2?1，求y?.
解 y??1x2?1)?x2lnxlnxxx2?1?x2?1?x2lnxxx2?1(x2?1)?x2?1?(3?x3.
x(x2?1)2(x2?1)24.求 y?xx?x?0?的导数.
解 对y?xx两端求自然对数，有lny?xlnx，对lny?xlnx两端求导
y?y?lnx?1 即
y??xx(lnx?1)
5.求函数y?arcsin2x?(arctanx)2的微分dy. 解 dy?d??arcsin2x???d?(arctanx)2?
?2txa2(arcsin2x)?(1?4x2)dx?2arc1?x2dnx
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n2 ?12arctanx????dx. ?2?(arcsin2x)?(1?4x2)?1?x??1.求极限limx?97?2x?5.
x?3解 limx?97?2x?5(2x?18)(x?3)2(x?3)6=lim?lim?. x?9(x?9)(7?2x?5)x?9(7?2x?5)5x?3x?x?2.求函数y???的导数.
?1?x?解 两边同时取对数，则有lny?xlnx，求导有 1?xy?x1?x1?ln?x??， y1?xx(1?x)2解得
x1??x??y???ln?????.
?1?x1?x??1?x?3.求极限lim(n?2?2n?1?n).
n??x解
lim(n?2?2n?1?n)?lim[(n?2?n?1)?(n?n?1)]
n??n?? ?lim(n?2?n?1)?lim(n?n?1)
n??n???limn??11?lim=0
n?2?n?1n??n?n?1ex4.设y?，求dy.
arctanxxx解 dy?arctanxde?edarctanx?2arctanxn??exarctanx?ex12x??(1?x)arctanx?1e2?dx. 1?xdx??arctan2x(1?x2)arctan2x5.求极限limn(n?8?n).
解 limn(n?8?n)=limn??n??8n?limn?8?nn??8=4 . 81??1n?x?1?1.求极限lim?? x??x?2??
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3x1111?1?[(1?)x]3xx)3x=lim(x)3x=lim解 lim(=e?3. xx??x??x??2221?1?[(1?)2]6xxx2x
x?0tan2x2x2x解 lim=limcos2xlim=1 .
x?0tan2xx?0x?0sin2x2.求极限lim8n2?2n?13.求极限lim 2n??2n?31212(8n?2n?1)8??228n?2n?1nnn 解 lim==limlimn??n??n??132n2?32(2n?3)2?n2n2218?lim?lim28?0?0n??nn??n?4 ==
32?02?lim2n??n24.求函数f(x)??解 f???lim?x?c?ax?b,2?3x,x?c,在c的左导数. 当a与b为何值时, 函数f(x)在c 可导. x?c.ax?b?(ac?b)?a
x?c2 若可导, 则f(x)在c点连续, 所以f(c?0)?f(c?0). 于是有3c?ac?b.
3x2?3c2?6c， 并且f??(c)?f??(c), 所以. f??(c)?limx?c?x?c?a?6c, 所以?时, 函数f(x)在c可导. 2b??3c.?1.求极限：lim(1?n??111)(1?)?(1?) 22223n解 注意
1?则
n??1(n?1)(n?1)? n2n21111*32*4(n?1)(n?1))(1?)?(1?)lim(*2?== 22222n??23n23nlim(1?2232?(n?1)2n(n?1)1n(n?1)1lim =lim22== n??2n??23?(n?1)2n22n222n2?12.求极限：lim2
n??n?n
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