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anonymous

  • 5 years ago

I don't see the reasoning in going straight from this limit to the answer is one.. someone help. (((1+n^(1/7)+n^(1/6))/(2-n+n^(4/3))) * (n^(7/6)) / 1

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  1. anonymous
    • 5 years ago
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    \[((1+ \sqrt[7]{n}+\sqrt[6]{n})\div(2-n+ \sqrt[3]{n ^{4}}))(\sqrt[6]{n ^{7}})\] Written out a little nicer

  2. anonymous
    • 5 years ago
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    Do u need to calculate limit?

  3. anonymous
    • 5 years ago
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    Well I'm doing a limit comparison test, and they went straight from this mess to the limit equals one.. I just need to know this limit doesn't equal 0 really..

  4. anonymous
    • 5 years ago
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    And it's as it goes to infinity

  5. anonymous
    • 5 years ago
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    check it out

  6. anonymous
    • 5 years ago
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    I already wolframmed it.. I clicked show steps and there were none, lol. but i'll check it out

  7. anonymous
    • 5 years ago
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    ok..

  8. anonymous
    • 5 years ago
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    yeah.. still no steps

  9. anonymous
    • 5 years ago
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    Lim comparison test, they took the function they were evaluating, they took another function that they know how it behaves, function a/function b take lim. Limit is 1is helpful info: lim>0 says function a and function b either both converge or both diverge.

  10. anonymous
    • 5 years ago
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    Yeah, I need help understanding why exactly that limit it one. It looks like there is too much going on to start L'hopitaling it.. And the top and both both have different powers when multiplied through by n^(7/6)

  11. anonymous
    • 5 years ago
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    For better readability: \[\frac{1 + n^{1/7} + n^{1/6}}{2-n +n^{4/3}} * \frac{n^{7/5}}{1}\]

  12. anonymous
    • 5 years ago
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    Is that right?

  13. anonymous
    • 5 years ago
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    Simplify, take lim of n^(highest exponent)/n^(highest exponent)

  14. anonymous
    • 5 years ago
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    yes except it's n^(7/6) / 1 at the end

  15. anonymous
    • 5 years ago
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    Ok, so distribute that n through we get \[\frac{n^{7/6} + n^{1/7 + 5/7} + n^{1/6 + 5/7}}{2-n + n^{4/3}}\]

  16. anonymous
    • 5 years ago
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    Err that's wrong.

  17. anonymous
    • 5 years ago
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    Got my fractions upside-down \[\frac{n^{7/6} + n^{1/7 + 7/6} + n^{1/6 + 7/6}}{2-n+n^{4/3}}\]

  18. anonymous
    • 5 years ago
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    lol.. thats just perfect isn't it

  19. anonymous
    • 5 years ago
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    oops...

  20. anonymous
    • 5 years ago
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    I was trying to multiply the exponents.. I see why it's one now

  21. anonymous
    • 5 years ago
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    So on top we have the largest power of n being \( n^{8/6}\)

  22. anonymous
    • 5 years ago
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    And on bottom, same thing.

  23. anonymous
    • 5 years ago
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    So yeah, 1.

  24. anonymous
    • 5 years ago
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    :P

  25. anonymous
    • 5 years ago
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    When you multiply powers of the same base you add their exponents.

  26. anonymous
    • 5 years ago
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    yeah....

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