Thursday, 29 May 2014

calculus - limlimitsxtoinftyfraclnxsqrtx,sinx



What a weird function.



I tried to find out: limxlnxxsinx




So, I can't use L'Hopital 'cause there's no actual limit in the denominator. It doesn't exist.



Then, I tried to use Heine's theorem and chose two sequences, but yet I got the same limit.



I believe it does not converge. How can I prove it?



Thanks


Answer



You can consider a sequence xn=πn2n. On this sequence your function will become unbounded while this sequence goes to infinity. Hence the limit does not exist.



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