loadinga+a+?a)5.设f(x)=(nx1x2xn1x且x≠0,ai>0,ai≠1,i=1,2,?,n,n>2x→+∞limf(x)求1)limf(x)2)limf(x)3)x→-∞x→0ln(a+?+a)-lnn解：1)limlnf(x)=limx→0x→0xx1xn=lna1?ann故
limf(x)=na1?anx→02）令M=max{a1,a2,?an}，我们有
a+?+aM≤nnxx1xn≤Mxn1xx即
Mn1xa+?+a≤(n1xx1)≤M而
limn=1，故limf(x)=max{a1,?,an}x→+∞x→+∞同理可得：
3)x→-∞limf(x)=min{a1,?,an}f(sinx+cosx)6.已知f(1)=0,f′(1)=2,求I=limx→0xtanx2f(1?sinx?cosx?1)?f(1)sinx?cosx?1I?lim?2x?0xtanxsinx?cosx?122?sinxcosx?1???f'(1)?lim?lim?x?0xtanxx?0x2???1?f?(1)(1?)?122(1?x)?e(1?ln(1?x))7.求极限limx?0x解:
I=lim2x2x→0(1?x)?e(1?ln(1?x))x2x222x2=lim(1?x)?e?eln(1?x)x→0x222x=lim(1?x)?e?limeln(1?x)x→0x?0xx22x2=lim(1?x)?e?ex→0x(1?x)?e2=lim?ex→0x2x2x22e?ln(1?x)?x?2=lim?ex→02x=lim?2e?e??2ln(1?x)?2xx→0??1????e2ln(1?x)?x2?2elim?ex?02x2