2楼: Originally posted by wurongjun at 2015-12-04 21:59:19
等价无穷小代换
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)→1/3

3楼: Originally posted by huangbl2014 at 2015-12-04 22:03:28
说了，不能用泰勒展开。...

4楼: Originally posted by wurongjun at 2015-12-04 22:06:52
你认为哪里用了泰勒展开?
tan(1/n)~1/n+1/n^3/3这个可以用罗比达法则证明啊!...

5楼: Originally posted by huangbl2014 at 2015-12-04 22:22:55
你换了一个名字而已，不要狡辩了。...

7楼: Originally posted by xmcchenchen at 2015-12-04 22:45:05
用归结原则，用洛必达几次就可以了

9楼: 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)
令t=(1/n),化简得到lim（tant-t)/t^3,洛必达法则化为lim((sect)^2-1)/t^3=lim((tant)^2)/(3t^2)=1/3