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,Âå±Ø´ï·¨Ôò»¯Îªlim((sect)^2-1)/t^3=lim((tant)^2)/(3t^2)=1/3

2Â¥: Originally posted by wurongjun at 2015-12-04 21:59:19
µÈ¼ÛÎÞÇîС´ú»»
tan(1/n)~1/n+1/n^3/3
n*tan(1/n)~1+1/n^2/3
ln(1+1/n^2/3)~1/n^2/3
n^2*ln(1+1/n^2/3)¡ú1/3

3Â¥: Originally posted by huangbl2014 at 2015-12-04 22:03:28
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