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Math Help - L'Hopital's rule question

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    [quote=curvature;59532][quote=kkoutsothodoros;59530] i made the mistake of not noticing that the denominator oscillates between infinity and - infinity.

    Note that the denominator only oscillates between e and 1/e. See the first graph above.
    he was talking about the graph he obtained by applying L'Hopital's rule i believe
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  2. #17
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    yes. i had created an inconsistency. i knew 1/e^sinx had no limit. i multiplied by 1 (x/x) and then applied L'Hopital's rule, INCORRECTLY, and then had an inconsistency because applying the rule was giving me that the limit was 0. it was this inconsistency i was trying to resolve.
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  3. #18
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    Quote Originally Posted by kkoutsothodoros View Post
    originally when i posted this thread i knew 1/e^sinx had no limit as x-->infinity. what i did was multiply top and bottom by x. i then used L'Hopital's rule and incorrectly showed it's limit was 0. i couldn't resolve the inconsistency. i assumed after taking derivative of top and bottom of x/xe^sinx that the derivative of the bottom (e^sinx+xcosxe^sinx) goes to infinity. it doesn't because it oscillates between positive and negative infininty. i was only seeing the positive!!
    The L'Hopital's rule can only be directly used for two types: type "0/0" and type "infinity/infinity"
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    The L'Hopital's rule can only be directly used for two types: type "0/0" and type "infinity/infinity"
    the latter condition was fulfilled when he multiplied by x/x
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  5. #20
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    yes, so that's why i applied it to x/xe^sinx. i believe the limit as x-->infininit for both x and xe^sinx is infininty.
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  6. #21
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    Quote Originally Posted by kkoutsothodoros View Post
    yes. i had created an inconsistency. i knew 1/e^sinx had no limit. i multiplied by 1 (x/x) and then applied L'Hopital's rule, INCORRECTLY, and then had an inconsistency because applying the rule was giving me that the limit was 0. it was this inconsistency i was trying to resolve.
    L'Hopital rule cannot be used for x/(x*exp(sinx)) since the denominator oscillates between e*x and x/e (see the graph below).
    Attached Thumbnails Attached Thumbnails L'Hopital's rule question-july06.gif  
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  7. #22
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    Quote Originally Posted by Jhevon View Post
    the latter condition was fulfilled when he multiplied by x/x
    Incorrect. see the last graph.
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  8. #23
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    yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.
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  9. #24
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    Quote Originally Posted by kkoutsothodoros View Post
    yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.
    Oh yes thanks
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  10. #25
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    all in all i think this was a good problem to get the wheels turning in my head!
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  11. #26
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    Quote Originally Posted by kkoutsothodoros View Post
    yes but if you consider the lower bound line in the graph it is going to infinity and therefore xe^sinx is going to infininty.
    However, after take derivatives, the limit of the new fraction dose not exist. So L'Hopital's rule cannot be used. The conditions of using the rule are not satisfied.
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  12. #27
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by curvature View Post
    However, after take derivatives, the limit of the new fraction dose not exist. So L'Hopital's rule cannot be used. The conditions of using the rule are not satisfied.
    The rule can be used, it just reaffirms our original thought. the limit still doesn't exist (which is what we originally thought but wanted verification)

    for some reason the poster wanted to convince himself that the infinite limit of e^{ \sin x} does not exist
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  13. #28
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    yes, if you draw one of those graphs again (could you post one?) for te denominator (e^sinx(1+xcosx)) after applying L'Hoptal's rule it will oscillate between infinity and -infinity so the limit doesn't exist (it is not infinity or -infininty) and so L'Hopital's rule, although used correctly, is not giving me an answer and so the incinsistency i was concerned with is resolved.
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  14. #29
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    or what Jhevon said!
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  15. #30
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    Quote Originally Posted by kkoutsothodoros View Post
    yes, if you draw one of those graphs again (could you post one?) for te denominator (e^sinx(1+xcosx)) after applying L'Hoptal's rule it will oscillate between infinity and -infinity
    Yes you are right.
    Attached Thumbnails Attached Thumbnails L'Hopital's rule question-july07.gif  
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