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Limit of a Sequence

  1. Aug 27, 2011 #1
    use the definition of a sequence to establish the limit
    lim([itex]\frac{2n}{n+1}[/itex])=2


    Let [itex]\epsilon[/itex]>0, then |[itex]\frac{2n}{n+1}[/itex]-2| <[itex]\epsilon[/itex]. Next we have that | [itex]\frac{2n-2n+-2}{n+1}[/itex]|= |[itex]\frac{-2}{n+1}[/itex]| <[itex]\frac{2}{n}[/itex]. So [itex]\exists[/itex] k[itex]\in[/itex][itex]N[/itex] such that [itex]\frac{2}{k}[/itex]<[itex]\epsilon[/itex]. When n[itex]\geq[/itex]k, we have [itex]\frac{2}{n}[/itex] < [itex]\frac{2}{k}[/itex] <[itex]\epsilon[/itex]. Therefore the limit is 2. BLOCK


    Is this the right way to do a limit proof?
     
  2. jcsd
  3. Aug 27, 2011 #2

    micromass

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    Looks good. Some remarks:

    2n+-2 is not something one writes. You'll have to write 2n+(-2) or simply 2n-2. I think you've just made a typo, but I wanted to make sure.

    Why does such a k exist?
     
  4. Aug 27, 2011 #3
    The Archimedean property right, is that something I should mention? Another question if we weren't given what the limit is equal to do you just make a guess and see if it right?
     
  5. Aug 27, 2011 #4

    micromass

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    Yes, you should certainly mention that!

    Yes, you need to make an educated guess by looking at the graph or perhaps by using properties of the limit. But I'm sure you will see some easy ways to calculate limits soon.
     
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