Friday, 25 October 2013

limits - How to compute limxtoinftyfrac2xx200?





I am trying to solve this limit:
limx2xx200



The first result is an indetermination of the kind but here applying L'Hopital would be too long, I do not see any substitution by means of equivalent infinitesimals possible and simplifying the limit neither.



How can I solve it? The solution must be . Thank you!


Answer



L'Hopital rule applied 200 times leads to
limx(log2)2002x200!=




Or you can take the logarithm of the limit



loglimx2xx200=limx(xlog2200logx)=limxx(log2200logxx)=



As the log of the limit is + the limit is +


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