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frx

  • 3 years ago

Let a>0 (a). Show using L'Hospitals rule that: \[\lim_{x \rightarrow 0} x ^{a}\ln(x)=0\] (b). By setting x=ln(t) in (a). show that: \[\lim_{x \rightarrow -\infty} |x|^{a} e ^{x}=0\] I need help with (b). I don't get the question, how do I show that (b). is true by setting x=ln(t) in (a). it doesn't make any sense to me.

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  1. sirm3d
    • 3 years ago
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    the proposed substitution

  2. sirm3d
    • 3 years ago
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    \[x=\ln t\] means that \[t=e^x\]

  3. frx
    • 3 years ago
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    I've got that but how can I use it?

  4. sirm3d
    • 3 years ago
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    as \[x \rightarrow -\infty, t \rightarrow ?\]

  5. frx
    • 3 years ago
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    Zero?

  6. frx
    • 3 years ago
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    e^-infinty = 0

  7. sirm3d
    • 3 years ago
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    correct

  8. sirm3d
    • 3 years ago
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    so we shift from

  9. sirm3d
    • 3 years ago
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    \[\lim_{x \rightarrow -\infty} \] to \[\lim_{t \rightarrow 0^+}\]

  10. sirm3d
    • 3 years ago
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    in terms of t, \[\left| x \right|^a = ?\]

  11. frx
    • 3 years ago
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    \[|\ln (t)|^{a}\] ?

  12. sirm3d
    • 3 years ago
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    that's right. and what about \[e^x\]

  13. frx
    • 3 years ago
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    We could set it as just t, right?

  14. sirm3d
    • 3 years ago
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    right

  15. frx
    • 3 years ago
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    \[\lim_{t \rightarrow 0} |\ln(t)|^{a}t\]

  16. frx
    • 3 years ago
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    Because t->0 shouldn't the whole expression be equal to 0 then?

  17. sirm3d
    • 3 years ago
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    not necessarily

  18. sirm3d
    • 3 years ago
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    what about \[\left| \ln t \right|^a\]

  19. frx
    • 3 years ago
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    \[\ln(t) \rightarrow -\infty\] but since it's the abolute value its just \[\infty \]

  20. sirm3d
    • 3 years ago
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    \[\left| \ln t \right|^a \rightarrow +\infty \]. you are correct

  21. sirm3d
    • 3 years ago
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    the transformed problem is an indeterminate form \[0*\infty \] which should be resolved either as \[\frac{ 0 }{ 0 }\] or \[\frac{ \infty }{ \infty }\]

  22. frx
    • 3 years ago
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    Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct

  23. frx
    • 3 years ago
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    Should I use the form \[\frac{ t }{ |\ln(t)|^{-a} }\] and then go for L'Hospital?

  24. sirm3d
    • 3 years ago
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    how about the other form?

  25. frx
    • 3 years ago
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    Just as it was, is it possible to use L'hostial on multiplication form?

  26. sirm3d
    • 3 years ago
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    lhopitals rule must be in quotient form

  27. frx
    • 3 years ago
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    Which form are you thinking about then?

  28. sirm3d
    • 3 years ago
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    i'm trying your quotient form

  29. frx
    • 3 years ago
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    \[-\frac{ a|\ln(t)|^{-a-1} }{ t }\] Can that be the correct derivation? If so it should give us the right answer \[\lim_{t \rightarrow 0}\frac{ 1 }{ -\frac{ a|\ln(t)|^{-a-1} }{ t } } = \frac{ 1 }{- \infty }=0\] Can this be right?

  30. sirm3d
    • 3 years ago
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    we'll look at the 2nd quotient form \[\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }\], noting that a>0

  31. sirm3d
    • 3 years ago
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    since t->0+, \[\left| \ln t \right|=-\ln t \]

  32. frx
    • 3 years ago
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    Right

  33. sirm3d
    • 3 years ago
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    the indeterminate form is inf/inf. so we'll apply l'hop's rule once

  34. sirm3d
    • 3 years ago
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    i hate getting disconnected in the middle of the equation

  35. frx
    • 3 years ago
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    \[\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }\]

  36. sirm3d
    • 3 years ago
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    right. when simplified becomes \[\frac{ a(-\ln t)^{a-1} }{ 1/t }\]

  37. sirm3d
    • 3 years ago
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    here's the catch to the problem

  38. sirm3d
    • 3 years ago
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    \[(-\ln t)^a \rightarrow +\infty \] for as long as the exponent a > 0.

  39. frx
    • 3 years ago
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    So it's still \[\frac{ \infty }{ \infty } \]

  40. sirm3d
    • 3 years ago
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    right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule

  41. frx
    • 3 years ago
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    \[\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }\]

  42. sirm3d
    • 3 years ago
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    after k applications of the rule, the indeterminate form is now \[a(a-1)(a-2)...(a-k+1)\frac{ (-\ln t)^{a-k} }{ 1/t }\] where a-k is negative

  43. sirm3d
    • 3 years ago
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    so that \[(-\ln t)^{a-k} \rightarrow 0\] because a-k is negative

  44. sirm3d
    • 3 years ago
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    the quotient form is now \[a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }\] and the problem is solved!

  45. frx
    • 3 years ago
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    Oh, that's clever, I get it

  46. frx
    • 3 years ago
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    Thank you so much for you're help I can't thank you enought this one has been on my mind for a couple of days now!

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