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## Homework Statement

What is the limit of [tex] { \frac{n+1}{2n} }[/tex] as

*n*--> oo. Prove your answer.

## The Attempt at a Solution

This is example from my book. Here is the problem:

*I am using the capital letter "

*E*" instead of "ε" in latex-code.

Intuitively, 1 is small relative to

*n*as

*n*gets large, so [tex] { \frac{n+1}{2n} } [/tex] behaves like [tex] { \frac{n}{2n} } [/tex], so we expect the limit to be 1/2 as

*n*--> oo. To prove this, we observe that

[tex] | S_n - L | = | { \frac{n+1}{2n} } - { \frac{1}{2} } | = | { \frac{1}{2n} } | = { \frac{1}{2n} } < E[/tex]

We observe that if n > 2/ε, then 0 < 1/2n < ε. Thus taking

*N*to be any integer > 2/ε, we have as required that

*n>N*, then [tex] |S_n - L | < E[/tex].

My question is, why didn't they choose

*N*to be any integer > 1/(2ε) ? Since:

[tex] { \frac{1}{2n} } < E[/tex] implies [tex] { \frac{1}{2E} } < n [/tex]

I see that 1/(2ε) < 1/ε < 2/ε <

*N*. Is there a reason why

*N*> 2/ε was chosen rather than

*N*> 1/(2ε)