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osanseviero Group Title

Determine if the next progression have a limit (demostrate it) and determine it

  • one year ago
  • one year ago

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  1. osanseviero Group Title
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    \[\left( -1^{n} \right)\left( \frac{ 5n+4 }{ 2n } \right)\]

    • one year ago
  2. osanseviero Group Title
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    Help, please?

    • one year ago
  3. osanseviero Group Title
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    :/

    • one year ago
  4. tkhunny Group Title
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    As n increases in the positive direction, \(\dfrac{5n+4}{2n}\) approaches \(\dfrac{5}{2}\). Follow with your mind as n increases. The little 4 on the end of the numerator becomes less and less significant. The terms do NOT approach zero.

    • one year ago
  5. osanseviero Group Title
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    oh...I think I see it now...but how can i demostrate if it is it's limit?

    • one year ago
  6. osanseviero Group Title
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    but there is also the -1^n

    • one year ago
  7. osanseviero Group Title
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    So it approaches 5/2 and -5/2 ?

    • one year ago
  8. tkhunny Group Title
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    The typical demonstration is a division by n. For n > 0, \(\dfrac{5n+4}{2n} = \dfrac{5 + \dfrac{4}{n}}{2}\). In this form, it is relatively obvious that the limit it 5/2 as n increases. The FIRST criterion for convergence is terms that approach ZERO. Nothing else will do. These terms do not approach zero, therefore, we do not care about the alternating sign. If the terms approach zero, THEN we'll worry about the sign.

    • one year ago
  9. osanseviero Group Title
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    what I mean is that there is a (-1^n) multiplying all of that...so -5/2 is also a limit

    • one year ago
  10. tkhunny Group Title
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    No, this is not a limit. Limits come alone, not in pairs. The terms, without the sign, approach 5/2. I may have stated that carelessly, before. The actual terms, including the sign, do not have a limit. They is oscillating.

    • one year ago
  11. osanseviero Group Title
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    Oh...okk

    • one year ago
  12. osanseviero Group Title
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    So for this there isnt a limit, neither \[\frac{ 1 }{ 2 },2^{2}, \frac{ 1 }{ 2^{3} }\]

    • one year ago
  13. tkhunny Group Title
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    Seriously? An alternating sign in the exponent? No. No limit.

    • one year ago
  14. osanseviero Group Title
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    I thought so, thanks :)

    • one year ago
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