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privetek Group Title

find the limit of (7n-8)/(2-sqrtn) as n goes to infinity

  • one year ago
  • one year ago

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  1. Dido525 Group Title
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    |dw:1352868502561:dw|

    • one year ago
  2. privetek Group Title
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    yes it is

    • one year ago
  3. Dido525 Group Title
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    |dw:1352869767197:dw|

    • one year ago
  4. Dido525 Group Title
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    Think about this logically. Which one approaches infinity faster?

    • one year ago
  5. privetek Group Title
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    thanks.. the answer is negative infinity, does that mean it diverges, right?

    • one year ago
  6. Dido525 Group Title
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    Actually... It's infinity. Not - Infinity.

    • one year ago
  7. Dido525 Group Title
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    Think. Which n term approaches infinity faster. In other words, does 7n and a square root approach infinity faster?

    • one year ago
  8. Dido525 Group Title
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    or a square root*

    • one year ago
  9. privetek Group Title
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    if you do L'Hopital's rule, you get 7/(-1/(2sqrtn)) = -14sqrtn limit of that is negative infinity

    • one year ago
  10. Dido525 Group Title
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    Lol. Okay fine We can use L'hopital's rule if you want.

    • one year ago
  11. Dido525 Group Title
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    Yes, then it is -Infinity. It diverges then.

    • one year ago
  12. privetek Group Title
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    :) Thanks

    • one year ago
  13. Dido525 Group Title
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    No problem :) .

    • one year ago
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