x2(1?xsinx?cosx)1?xsinx?cosx)?lim解 lim ?limx?0x?01?xsin1?cosxxsinxx?01?xsinx?cosxx?cosx?2xx2x2?1?14?. 1?132
x?sin2x.
x?0x?sin2xsin2xsin2x1?1?2x?sin2xx?lim2x?1?2??1. 解 lim?limx?0x?sin2xx?0sin2xx?0sin2x1?231?1?2x2x
例18 (E07) 计算 lim?1?例19 (E08) 求 lim?1??n???n?n?3. 1?????1?n??n?1?解 lim?1??n???n?n?3???lim??1?n?????1??n?3??1??1???lim?1????1???e?1?e.
n???n??n???n3
1/x例20 (E09) 求 lim(1?2x).
x?0解
1lim(1?2x)xx?01????lim?(1?2x)2x?x?0?????2?e?2.
?k?例21 (E10) 求lim?1??.
x???x?xx????kkkk?k?????k解 lim?1???lim??1?????lim?1????e.
x??x????x???x???x??x??????xkkx?1?特别地，当k??1时，有lim?1???e?1.
x???x?
?3?x?例22 (E11) 求 lim??.
x???2?x??3?x?解 lim??x???2?x?2xxx?2?2?????1??1??lim??1?? ???lim??1??x????x??x?2x?2??????????x?2?4??1???1?2?lim??1????1???e. x????x?2??x?2????222x2x
?x2??. 例23 求 lim?x???x2?1???xx???x2?11??????lim解 lim??1?2??lim??1?2x???x2?1?x????x???x?1x?1?????xxx2?1?x?12????e0?1.
x1/x例24 计算 lim(e?x).
x?01(ex?解 limx?01x)x?1?lim(ex)x?1?x?0?e?x?xx???elim1???x?x?0?ex???e??1?xx?ex?2??e?e?e. ??
tan2x. 例25 求极限 lim(tanx)x??/4解 令t?tanx?1,则tanx?t?1,当x??4时,t?0,又
tan2x?2(t?1)2tanx12(t?1)? ???22tt?21?tanx1?(t?1)12(t?1)??lim(1?t)tt?2t?01?2(t?1)lim[(1?t)t]t?2t?0故lim(tanx)tan2x?x???1?[lim(1?t)t]t?0limt?0?2(t?1)t?2?e?1.
4
连续复利
例26 (E12) 小孩出生之后，父母拿出P元作为初始投资，希望到孩子20岁生日时增长到100000元，如果投资按8%连续复利，计算初始投资应该是多少？
解 利用公式S?Pe，求P. 现有方程
rt100000?Pe0.08?20
由此得到
e P?100000?1.6?20189.65
于是，父母现在必须存储20189.65元，到孩子20岁生日时才能增长到100000元.
计算现值可以理解成从未来值返回到现值的指数衰退.
一般地，t年后金额S的现值P, 可以通过解下列关于P的方程得到
S?Pekt，P?
P?kt?Pe. ekt课堂练习
1. 求极限 limtanx?sinx.
x?0x2sinx2. 求极限lim
x???1(3x?9x)x.