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Limits and Sequencing


Date: 5/3/96 at 10:15:15
From: Anonymous
Subject: Limits and sequencing

This is an approaching limit question, from a maths limits question
this one's supposed to be pretty hard...

Find the limit a, and find an expression for the number N(epsilon) 
such that

|a(n) - a| < epsilon for all n > N(epsilon)
a(n) = 1/n sin^2 (n.pi/4)

where a(n) is where the n is a subscript


Date: 10/17/96 at 8:5:49
From: Doctor Jerry
Subject: Re: Limits and sequencing

If a(n) = (1/n)*sin^2(pi*n/4), which is what I think you may intend, 
the question is not very hard (although it looks as if it might be).

The factor sin^2(pi*n/4) always lies between 0 and 1.  Consequently, 

0<=a(n)<=1/n.

Since 1/n goes to zero as n becomes infinite, so does a(n).  
Specifically, since 1/n < epsilon when n > 1/epsilon, we may take 
N(epsilon)  the first positive integer greater than 1/epsilon. 

When n > N, |a(n) - 0| < epsilon.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis

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