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LaPlace Transform question #1

  1. Apr 13, 2013 #1
    1. The limit as b approaches infinity always shows up as undefined on my calc so I don't know what to put for that section of the work.

    2. What pat of the work is supposed to need L'hopital's rule? The integration?

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  2. jcsd
  3. Apr 13, 2013 #2

    Dick

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    Same type of answer as with the other one. In e^(b(9-s)), 9-s is negative. That will go to zero as b->infinity. So will b*e^(b(9-s)) and b^2*e^(b(9-s)).
     
  4. Apr 13, 2013 #3
    What about the 2nd question?
     
  5. Apr 13, 2013 #4

    Dick

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    About l'Hopitals's? You can use it to prove something like b*exp(-kb) must go to zero as b->infinity if k>0. Write it as b/e^(kb) and use l'Hopital.
     
  6. Apr 13, 2013 #5
    How does it specifically apply in this case? What am I supposed to need it for in this problem?
     
  7. Apr 13, 2013 #6

    Dick

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    You want to show things like -b^2*e^(b(9-s))/(9-s) actually do go to zero when s>9 and b goes to infinity.
     
  8. Apr 13, 2013 #7
    Ok, so it's supposed to be needed when evaluating the limits of each term correct?
     
  9. Apr 13, 2013 #8

    Dick

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    Yes, the parts you wrote '?' over. You'll want to show they actually do go to zero if s>9.
     
  10. Apr 14, 2013 #9
    1CnPNon.jpg

    Did I do the work right? I get 0 in the denominator.. what's wrong here?
     
  11. Apr 14, 2013 #10

    Dick

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    You can't use l'Hopital's rule on an expression unless it's indeterminate. Write it so it has the form 0/0 or infinity/infinity. Try b^2/e^b. That's infinity/infinity.
     
  12. Apr 14, 2013 #11
    ok. So for that specific term that I tried to do, the limit looks like it would instead be infinity right? Am I allowed to do the limit of just -b^2 over s-9 by itself, along with each other term by itself, and the e^b(9-s) by itself, or do I have to expand it out and then do each term like I started to do and do the limit that way?
     
  13. Apr 14, 2013 #12

    Dick

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    No, it's not infinity. It's zero. Limit b^2/e^b is infinity/infinity. Do l'Hopital. 2b/e^b. Still infinity/infinity. Do l'Hopital once more. Now it's not infinity/infinity. What is it?
     
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