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privetek

  • 3 years ago

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

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  1. Dido525
    • 3 years ago
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    |dw:1352868502561:dw|

  2. privetek
    • 3 years ago
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    yes it is

  3. Dido525
    • 3 years ago
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    |dw:1352869767197:dw|

  4. Dido525
    • 3 years ago
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    Think about this logically. Which one approaches infinity faster?

  5. privetek
    • 3 years ago
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    thanks.. the answer is negative infinity, does that mean it diverges, right?

  6. Dido525
    • 3 years ago
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    Actually... It's infinity. Not - Infinity.

  7. Dido525
    • 3 years ago
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    Think. Which n term approaches infinity faster. In other words, does 7n and a square root approach infinity faster?

  8. Dido525
    • 3 years ago
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    or a square root*

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

  10. Dido525
    • 3 years ago
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    Lol. Okay fine We can use L'hopital's rule if you want.

  11. Dido525
    • 3 years ago
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    Yes, then it is -Infinity. It diverges then.

  12. privetek
    • 3 years ago
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    :) Thanks

  13. Dido525
    • 3 years ago
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    No problem :) .

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