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m.auld64 Group Title

can someone please explain to me why the answer to the following question is -infinity, infinity

  • 2 years ago
  • 2 years ago

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  1. jaersyn Group Title
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    ticktock

    • 2 years ago
  2. m.auld64 Group Title
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    \[\lim_{x \rightarrow -infinity} (\ln(cbrt(x)))/sinx\]

    • 2 years ago
  3. m.auld64 Group Title
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    cbrt = cube root

    • 2 years ago
  4. jaersyn Group Title
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    lnx^(1/3)/sinx = (1/3)lnx / sinx ? i thinkk..

    • 2 years ago
  5. m.auld64 Group Title
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    actually apparently -infinity, infinity is not the correct answer.... wolfram lied so if you help me come to the correct answer

    • 2 years ago
  6. satellite73 Group Title
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    ?? \[\lim_{x\to -\infty}\frac{\ln(\sqrt[3]{x})}{\sin(x)}\]

    • 2 years ago
  7. m.auld64 Group Title
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    ya thats it

    • 2 years ago
  8. satellite73 Group Title
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    makes no sense since you cannot take the log of a negative number

    • 2 years ago
  9. myko Group Title
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    i think should be 0

    • 2 years ago
  10. satellite73 Group Title
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    so there is no limit, as the numerator is undefined if \(x\leq 0\)

    • 2 years ago
  11. m.auld64 Group Title
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    oh crap it is \[\lim_{x \rightarrow +infinity}\] sorry....

    • 2 years ago
  12. satellite73 Group Title
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    \(\sin(x)\) has no limit either as it is periodic and takes on all values between -1 and 1 infinitely often

    • 2 years ago
  13. satellite73 Group Title
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    well still no limit

    • 2 years ago
  14. satellite73 Group Title
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    numerator is going to infinity, but denominator is not going to any specific number as i wrote above.

    • 2 years ago
  15. satellite73 Group Title
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    in other words it swings wildly between large positive and large negative values as sine varies between -1 and 1

    • 2 years ago
  16. m.auld64 Group Title
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    so ultimately its undefined

    • 2 years ago
  17. satellite73 Group Title
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    yes for sure undefined

    • 2 years ago
  18. m.auld64 Group Title
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    so with that being said, anytime sinx is in my denomonator i know that the function must be undefined?

    • 2 years ago
  19. myko Group Title
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    hmm

    • 2 years ago
  20. jaersyn Group Title
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    no

    • 2 years ago
  21. satellite73 Group Title
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    well no, the numerator could go to zero, so the whole thing could go to zero

    • 2 years ago
  22. satellite73 Group Title
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    \[\lim_{x\to\infty}\frac{e^{-x}}{\sin(x)}\] for example would be zero

    • 2 years ago
  23. m.auld64 Group Title
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    so are there any good tips and tricks you could tell me to help me understand this sort of material

    • 2 years ago
  24. satellite73 Group Title
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    no tricks really, you have to make sure you know what it going on, and don't for example us l'hopital's rule unless it is applicable, i.e. unless you are in indeterminate form, which you are not here. in your example you do not have \(\frac{\infty}{\infty}\) and so you cannot use l\hopital and get zero for an answer!

    • 2 years ago
  25. m.auld64 Group Title
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    see im not sure how to determine whether i am in that form or not

    • 2 years ago
  26. satellite73 Group Title
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    plug in the number and check or imagine what happens as x gets large if you are going to infinity

    • 2 years ago
  27. satellite73 Group Title
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    in your example as x gets large so does \(\sqrt[3]{x}\) and therefore so does \(\ln(\sqrt[3]{x})\) but the prolem is that sine does not get large and does not have a limit

    • 2 years ago
  28. satellite73 Group Title
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    so basically what i am saying is that in this problem there is no trick or gimmick. you just have to imagine what happens as x gets bigger and bigger

    • 2 years ago
  29. m.auld64 Group Title
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    ok thanks!

    • 2 years ago
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