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osanseviero

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

  • 5 months ago
  • 5 months ago

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

    • 5 months ago
  2. osanseviero
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    Help, please?

    • 5 months ago
  3. osanseviero
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    :/

    • 5 months ago
  4. tkhunny
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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.

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

    • 5 months ago
  6. osanseviero
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    but there is also the -1^n

    • 5 months ago
  7. osanseviero
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    So it approaches 5/2 and -5/2 ?

    • 5 months ago
  8. tkhunny
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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.

    • 5 months ago
  9. osanseviero
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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

    • 5 months ago
  10. tkhunny
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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.

    • 5 months ago
  11. osanseviero
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    Oh...okk

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

    • 5 months ago
  13. tkhunny
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    Seriously? An alternating sign in the exponent? No. No limit.

    • 5 months ago
  14. osanseviero
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    I thought so, thanks :)

    • 5 months ago
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