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krypton

  • 3 years ago

y=lmx->inf (xe^x)/(x^2+2x-1) use l hopitals rule

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  1. zepdrix
    • 3 years ago
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    Before using L'Hopital's Rule, verify that the limit is of an indeterminate form. This limit appears to be of the indeterminate form: \[\frac{ \infty }{ \infty }\] So we can apply L'Hopital's Rule. Do you need help applying the rule? :)

  2. krypton
    • 3 years ago
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    yeah i do pls

  3. krypton
    • 3 years ago
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    beacuse after applying the rule, i syill got inf/inf

  4. zepdrix
    • 3 years ago
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    |dw:1349037732010:dw| Still getting Inf/Inf? Good that is still an indeterminate form, which means you can apply L'Hop again! :)

  5. krypton
    • 3 years ago
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    so how do i apply it again?

  6. krypton
    • 3 years ago
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    after the 1st derivative, i got e^x(1+x)/(2x+2) then i plugged in inf and got inf/inf

  7. zepdrix
    • 3 years ago
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    |dw:1349037925108:dw| Yah you need to apply L'Hop from that spot again :)

  8. zepdrix
    • 3 years ago
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    The important thing to remember is that - You can only apply L'Hop rule a second time (or even third time, etc.. ) if it is STILL of an indeterminate form after the first application of L'Hop. Still confused? :O

  9. krypton
    • 3 years ago
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    yeah i now understand thanks alot :)

  10. zepdrix
    • 3 years ago
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    yay \c:/

  11. krypton
    • 3 years ago
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    hey so the answer will be?

  12. krypton
    • 3 years ago
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    infinity?

  13. zepdrix
    • 3 years ago
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    So it looks like it simplifies to:\[\frac{ \infty }{ 2 }\] Which, yes is simply Infinity. Good job! :)

  14. krypton
    • 3 years ago
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    hi i need a bit help with one other scenario as i wasnt in the class when the whole topic was thought how about this y=limx->0^+ x^2x

  15. krypton
    • 3 years ago
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    thats x approches 0 from the right

  16. zepdrix
    • 3 years ago
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    |dw:1349039331256:dw|

  17. zepdrix
    • 3 years ago
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    |dw:1349039504362:dw|

  18. zepdrix
    • 3 years ago
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    |dw:1349039633140:dw|

  19. zepdrix
    • 3 years ago
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    |dw:1349039718975:dw|

  20. zepdrix
    • 3 years ago
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    |dw:1349039844810:dw|

  21. zepdrix
    • 3 years ago
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    |dw:1349039949728:dw| Get stuck on any part of that? I tried to be detailed, since you missed the concept in class D': It's a little tricky.

  22. krypton
    • 3 years ago
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    1?

  23. zepdrix
    • 3 years ago
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    Yay \:D/

  24. krypton
    • 3 years ago
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    does this math change if the say x approaches 0 from the left??

  25. krypton
    • 3 years ago
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    0^-

  26. zepdrix
    • 3 years ago
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    Yes. The reason they made that distinction, is because the Natural log is only defined for values greater than 0. If we were approaching 0 from the left, it wouldn't make any sense, because that value isn't in the domain of lnx. So we are forced to approach from positive values! :)

  27. zepdrix
    • 3 years ago
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    |dw:1349040248054:dw|

  28. krypton
    • 3 years ago
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    oh! ok thanks alot for your help.so this as far as l'hopitals rule goes, no other thing? if is inderterminant u keep using the rule and taking the 2nd derivate

  29. zepdrix
    • 3 years ago
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    Yes repeat as often as you are able to. :) Important note: You need the CORRECT indeterminate form to apply L'Hop. If I remember correctly you need one of these two forms:\[\frac{ 0 }{ 0 },\frac{ \infty }{ \infty }\] In this problem we just did, we had the indeterminate form: \[\infty*0\]So we had to move things around before we could apply L'Hop :)

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