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Introduction to limits

Question

 Let fp(α)=eiαp2e2iαp2.eiαp, pN where i=1 , then find the value of limnfn(π)

Difficult
Solution

fp(α)=eiαp2(1+2++p)=eiα21+1plimnfn(π)=limneiπ21+1n=eiπ2limnfn(π)=eiπ2=1 



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