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osanseviero

  • one year ago

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

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

  2. osanseviero
    • one year ago
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    Help, please?

  3. osanseviero
    • one year ago
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    :/

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

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

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

  8. tkhunny
    • one year ago
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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.

  9. osanseviero
    • one year ago
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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

  10. tkhunny
    • one year ago
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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.

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

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

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

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

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