Tuesday, 22 May 2018

calculus - Calculating this limit without l'Hospital/Taylor/derivatives



Let's consider an=n2log(cos1n)




It's easy to calculate limnan

by using l'Hospital/Taylor. But how to do it without using anything that involves derivatives? (Pure sequences!)


Answer



I'll use two facts 1. limx0sinx/x=1. 2. limx0(1+ax+o(x))1/x=ea for any constant a.



From 1. we get, as x0,



1cosxx2=11+cosx1cos2xx2=11+cosxsin2xx21212=12.



This shows cosx=1(1/2)x2+o(x2). Therefore




[cos(1/n)]n2=[1+(1/2)/n2+o(1/n2)]n2e1/2,



where we have used 2. above. Now apply ln to see the desired limit is 1/2.


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