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ParthKohli

How is \[ \lim_{x \to \infty} 1 - 5x^3 = -5\]?

  • one year ago
  • one year ago

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  1. amistre64
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    is x and z the same?

    • one year ago
  2. sami-21
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    ok divide by x^2

    • one year ago
  3. sami-21
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    after division by x^2 limit x->infinity 1/x^2-5 can you do it now?

    • one year ago
  4. amistre64
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    cubics dont converge

    • one year ago
  5. ParthKohli
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    Hmm, \(0 - 5\) if we do it individually.

    • one year ago
  6. ParthKohli
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    I got it!

    • one year ago
  7. sami-21
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    ok .

    • one year ago
  8. ParthKohli
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    But how are we allowed to divide by \(x^2\)?

    • one year ago
  9. amistre64
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    5x^3/x^2 = 5x tho ....

    • one year ago
  10. ParthKohli
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    And how did you get \(5x^3 \div x^2 = 5\)? It's 5x.

    • one year ago
  11. ParthKohli
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    Yeah

    • one year ago
  12. ParthKohli
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    It's the same if we divide by \(x^3\) anyway, just that the thing becomes 1/x^3

    • one year ago
  13. sami-21
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    does not matter you divided both if you take x^2 as LCM you get original back.

    • one year ago
  14. sami-21
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    no it wouldn't be same. always divide by highest power available in the question . here it is x^2

    • one year ago
  15. ParthKohli
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    Aha! Thank you! But I still don't get how we are allowed to divide both sides by x^3.

    • one year ago
  16. sami-21
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    sorry it is x^3

    • one year ago
  17. ParthKohli
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    \[ \lim_{x \to \infty} x^3({1 \over x^3} - 5)\] \[ x^3 \lim_{x \to \infty} -5 \]

    • one year ago
  18. ParthKohli
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    I think that you also factor the \(x^3\) out.

    • one year ago
  19. ParthKohli
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    Oh! Wait!

    • one year ago
  20. ParthKohli
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    \[ \lim_{x \to \infty}x^5 \times \lim_{x \to \infty}{1 \over x^3} - 5\]

    • one year ago
  21. ParthKohli
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    x^3*

    • one year ago
  22. ParthKohli
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    But isn't that the indeterminate form?

    • one year ago
  23. amistre64
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    why are you trying to makes sense of this? a cubic does not converge

    • one year ago
  24. ParthKohli
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    But it is given that the solution is -5.

    • one year ago
  25. amistre64
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    and i say the sky has floating cotton candy for clouds; does it make it true?

    • one year ago
  26. sami-21
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    no you cannot take it out always.just divide each term because if you take common then you will get x^3also then it will always gives you infinity.

    • one year ago
  27. amistre64
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    http://www.wolframalpha.com/input/?i=limit%28x+to+inf%29+%281-5x%5E3%29

    • one year ago
  28. ParthKohli
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    Well, may I know how you are allowed to divide by \(x^3\)? (I know that you divide the variable with the exponent as the highest power).

    • one year ago
  29. apoorvk
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    Uh oh. this should pretty simply be equal to negative infinity. Logically speaking.

    • one year ago
  30. ParthKohli
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    The original question is: \[ \lim_{z \to 0} {4z^2 + z^6 \over 1- z^3} \]

    • one year ago
  31. amistre64
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    if the original is that, why would you ask this other thing as a post?

    • one year ago
  32. ParthKohli
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    Infinity, not 0.

    • one year ago
  33. sami-21
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    @amistre64 you are right .it does not converges . this division rule works only when you have fractions .rational functions @ParthKohli you should take common x^3. for example

    • one year ago
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  34. ParthKohli
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    Because they showed something like \[ \lim_{z \to \infty} {4z^2 + z^6 \over 1 - z^3} = {-\infty \over -5} \]

    • one year ago
  35. sami-21
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    the answer to your question is -infinity @ParthKohli

    • one year ago
  36. ParthKohli
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    Hmm, I was confused as to why it was -5 and not -infinity :P

    • one year ago
  37. ParthKohli
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    I mean... the question that I originally asked is supposed to be -infinity, right?

    • one year ago
  38. sami-21
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    yes right

    • one year ago
  39. apoorvk
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    Aye aye!

    • one year ago
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