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anonymous

  • 5 years ago

lim xln(x^2) x->oo

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  1. anonymous
    • 5 years ago
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    \[\lim_{x\rightarrow\infty}x\ln (x^2)\]

  2. anonymous
    • 5 years ago
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    or is it \[\lim_{x\rightarrow 0}x\ln (x^2)\]

  3. anonymous
    • 5 years ago
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    Yes that's correct. Need help! Not sure what I'm supposed to do next.

  4. anonymous
    • 5 years ago
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    No limit as x approaches infinity.

  5. anonymous
    • 5 years ago
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    if x is approaching infinity then xln(x^2) is approaching infinity

  6. anonymous
    • 5 years ago
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    not much to finding the limit there

  7. anonymous
    • 5 years ago
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    No I'm supposed to use L'hospital's rule or something like that.

  8. anonymous
    • 5 years ago
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    if x was approaching zero then you may have to but as x goes to infinity that function goes to infinity, i.e. the form you get \[\infty\infty\] is not indeterminate

  9. anonymous
    • 5 years ago
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    Okay how about lim

  10. anonymous
    • 5 years ago
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    i.e. a really large positive number multiplied by another really large number is just another really large positive number

  11. anonymous
    • 5 years ago
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    Sorry how bout lim xln(1-2/x) x->oo

  12. anonymous
    • 5 years ago
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    yes, as x approaches infinity the function is ever increasing to infinity, thus that is an infinite limt

  13. anonymous
    • 5 years ago
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    This is probably just a trick question thrown into your problem set where you are using L'Hopital's rule to see if they can catch you off guard

  14. anonymous
    • 5 years ago
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    sorry, i did not notice you changed the function

  15. anonymous
    • 5 years ago
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    \[\lim_{x\rightarrow\infty} x\ln(1-\frac{2}{x})\] so first you notice that you get the indeterminate form \[\infty 0\]

  16. anonymous
    • 5 years ago
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    and you need either \[\frac{0}{0}\operatorname{or}\frac{\infty}{\infty}\] to use L'Hopital

  17. anonymous
    • 5 years ago
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    so rewrite the function in the quivelent form \[\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{2}{x})}{\frac{1}{x}}\] and notice now the form is correct for L'Hopital

  18. anonymous
    • 5 years ago
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    after one iteration of L'Hopital's rule I get the limit is -2

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