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2Â¥: Originally posted by wurongjun at 2015-12-04 21:59:19
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tan(1/n)~1/n+1/n^3/3
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wurongjun

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4Â¥: Originally posted by wurongjun at 2015-12-04 22:06:52
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huangbl2014

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5Â¥: Originally posted by huangbl2014 at 2015-12-04 22:22:55
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xmcchenchen

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huangbl2014

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7Â¥: Originally posted by xmcchenchen at 2015-12-04 22:45:05
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weiweiai

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n^2*ln(n*tan(1/n))=n^2*(n*tan(1/n)-1)=n^2*(tan(1/n)-(1/n))/(1/n)
Áît=(1/n),»¯¼òµÃµ½lim£¨tant-t)/t^3,Â屨´ï·¨Ôò»¯Îªlim((sect)^2-1)/t^3=lim((tant)^2)/(3t^2)=1/3
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huangbl2014

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ÒýÓûØÌû:
9Â¥: Originally posted by weiweiai at 2015-12-04 23:02:30
n^2*ln(n*tan(1/n))=n^2*(n*tan(1/n)-1)=n^2*(tan(1/n)-(1/n))/(1/n)
Áît=(1/n),»¯¼òµÃµ½lim£¨tant-t)/t^3,Â屨´ï·¨Ôò»¯Îªlim((sect)^2-1)/t^3=lim((tant)^2)/(3t^2)=1/3

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