Ashley P.

asked • 10/10/20

Limit of a Sequence- Epsilon Delta Definition

Let an be a sequence of real numbers such that lim n-->infinity (an)=3.

By directly using the definition of the limit of a sequence, show that lim n-->infinity(2an/(2an+3))=2


Kevin S.

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You can't prove that because it's false. Is it supposed to be equal to 2/3?
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10/10/20

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Sebastian M. answered • 10/10/20

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Ashley you wrote the question out wrong. Here's a counter example:


Consider the sequence (an) such that an=3 for all n. Then (an)→3.


Then consider the sequence (bn) such that bn= 2an/(2an+3) = 2(3)/(2(3)+3) = 6/9 for all n


Which implies that (bn)→6/9≠2



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Sebastian M.

If you meant that (an)->-3 then the problem makes sense. Here's the proof: PF. Let ε>0. There exists N such that if n>N then |an-(-3)|=|an+3|<3ε/(2+2ε). We must show that this implies that |[2an/(2an+3)]-2|<ε. Through some simple algebraic manipulation you can show that |[2an/(2an+3)]-2| = 2|[an/(2an+3)]-1| = 2|(an+3)/(2an+3)|. Next consider that the denominator |2an+3|=|(2an+3)+3-3|=|2(an+3)-3|>=3-2|an+3| by the triangle inequality. Therefore 1/|2an+3|<=1/(3-2|an+3|). Now since |an+3|<3ε/(2+2ε) then 1/(3-2|an+3|)<1/[3-2(3ε/(2+2ε))]. Therefore, 2|an+3|/|2an+3|< 2*3ε/(2+2ε)/[3-2(3ε/(2+2ε))]. Through some more manipulation you can show that this expression is equal to 3ε/(1+ε)*(1+ε)/3 = ε. Therefore for ε>0 if n>N then [2an/(2an+3)]-2|< ε. Thus (2an/(2an+3))->2 as n->infinity.
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10/11/20

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