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anonymous

  • one year ago

Need Help! Find the limit of the limit below

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  1. anonymous
    • one year ago
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    |dw:1439230879897:dw|

  2. hartnn
    • one year ago
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    when v->4+ means v is very very near to 4, but greater than 4. agreed?

  3. anonymous
    • one year ago
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    Yep

  4. hartnn
    • one year ago
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    so, 4-v will be positive or negative?

  5. anonymous
    • one year ago
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    negative?

  6. hartnn
    • one year ago
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    thats right :) and since the denominator is having a modulus, it will be positive only. so our final answer will be negative!

  7. anonymous
    • one year ago
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    |dw:1439231347038:dw|

  8. anonymous
    • one year ago
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    I understand by writing things out..

  9. hartnn
    • one year ago
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    i meant |4-v| is always positive and 4-v is negative

  10. anonymous
    • one year ago
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    So how would I solve it?

  11. hartnn
    • one year ago
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    4-v from numerator and denominator would cancel out, so the answer would be either +1 or -1 the numerator is negative, and denominator is positive, so the overall answer is negative, that is -1.

  12. anonymous
    • one year ago
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    Are you sure it's that simple?

  13. hartnn
    • one year ago
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    yep!

  14. anonymous
    • one year ago
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    |dw:1439231746924:dw|Is that how the work would look like?

  15. hartnn
    • one year ago
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    ok, lets make some steps, since v is very near to 4, but greater than 4, lets take v = 4 + \(\delta \) when \(v \to 4^+\), \(\delta \to 0\) |dw:1439231931131:dw| try to simplify that ^^

  16. anonymous
    • one year ago
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    1?

  17. hartnn
    • one year ago
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    the numerator simplifies to \(-\delta\) and the denominator simplifies to \(|-\delta | = \delta\) isn't it ?

  18. hartnn
    • one year ago
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    |dw:1439232227653:dw|

  19. anonymous
    • one year ago
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    I see it now! thanks for the clear work

  20. hartnn
    • one year ago
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    welcome ^_^

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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