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m.auld64

  • 2 years ago

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

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  1. jaersyn
    • 2 years ago
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    ticktock

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

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

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

  5. m.auld64
    • 2 years ago
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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

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

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

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

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

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

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

  12. satellite73
    • 2 years ago
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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

  13. satellite73
    • 2 years ago
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    well still no limit

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

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

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

  17. satellite73
    • 2 years ago
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    yes for sure undefined

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

  19. myko
    • 2 years ago
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    hmm

  20. jaersyn
    • 2 years ago
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    no

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

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

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

  24. satellite73
    • 2 years ago
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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!

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

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

  27. satellite73
    • 2 years ago
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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

  28. satellite73
    • 2 years ago
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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

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

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