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求极限
limn→∞2n2−2+2+⋯2√−−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√.
解.
2√=2cosπ4=2cosπ222+2√−−−−−−√=2(1+cosπ22)−−−−−−−−−−−√=2cosπ23⋯2−2+2+⋯2√−−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√=2cosπ2n+1.
故
====limn→∞2n2−2+2+⋯2√−−−−−−−−√−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√limn→∞2n2−2cosπ2n+1−−−−−−−−−−−√limn→∞2n⋅2sinπ2n+2limn→∞2n+1π2n+2π2.□