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anonymous

  • 5 years ago

Let f=R->R be such that f(1)=3 and f '(1)=6 , then lim(x tends to zero) [(f(1+x)/f(1)]^(1/x) equals to ----------

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  1. anonymous
    • 5 years ago
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    Let f=R->R be such that f(1)=3 and f '(1)=6 , then \[\lim_{x \rightarrow 0}[(f(1+x)/f(1)]^{1/x}\] equals to ----------

  2. anonymous
    • 5 years ago
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    infinity, I guess :) since the power will be infinity, anything to the power of infinity is infinity :)

  3. anonymous
    • 5 years ago
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    or you can solve it using e and ln too

  4. anonymous
    • 5 years ago
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    wait....that's not the case here, what's the function?

  5. nowhereman
    • 5 years ago
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    you don't need the function Just write it as \[\exp\left(\frac {\ln \frac{ f(1+x)}{f(1)}}{x}\right)\] and you can solve it with l'hopital.

  6. anonymous
    • 5 years ago
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    Could you please explain how to come to that step?

  7. nikvist
    • 5 years ago
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    limes doesn't exist, because \[\lim_{x\rightarrow 0-}[f(1+x)/f(1)]^{1/x}\neq\lim_{x\rightarrow 0+}[f(1+x)/f(1)]^{1/x}\]

  8. nikvist
    • 5 years ago
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    wrong, my last post is not true

  9. nowhereman
    • 5 years ago
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    I just put exp(ln(.)) around the function and used logarithm rules.

  10. anonymous
    • 5 years ago
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    Okay, I will certainly look at it, though I have to leave again. If I can get a chance to post tonight, I will.

  11. anonymous
    • 5 years ago
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    That alright?

  12. anonymous
    • 5 years ago
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    Ok, al right

  13. nowhereman
    • 5 years ago
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    What is it you don't understand?

  14. anonymous
    • 5 years ago
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    Its ok, I understand what you said

  15. anonymous
    • 5 years ago
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    I have something else to ask

  16. anonymous
    • 5 years ago
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    I just wish to ask, what are the things to consider while solving problems on limit and continuity

  17. nowhereman
    • 5 years ago
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    I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.

  18. anonymous
    • 5 years ago
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    No no, you don't have to go to that. I am in high school

  19. anonymous
    • 5 years ago
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    AYCC is a homework tutoring marketplace. If you're a student seeking smart solutions to your difficult homework assignments or someone who’d love to pass on your pearls of wisdom for a specific topic or a question and earn additional income, join www.aceyourcollegeclasses.com

  20. anonymous
    • 5 years ago
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    iam, I can respond now if you're around.

  21. anonymous
    • 5 years ago
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    Yes I am online

  22. anonymous
    • 5 years ago
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    Could you please come here https://docs.google.com/document/d/1-_5IUkf1O4EFkBw-kDEZV1T8THk4lzrpmQEzVKtfHHo/edit?hl=en_GB#

  23. anonymous
    • 5 years ago
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    \[y=\left( \frac{f(x+1)}{f(1)} \right)^{1/x} \rightarrow \log y = \log \left( \frac{f(x+1)}{f(1)} \right)^{1/x}\]That is\[\log y =\frac{1}{x}\log \left( \frac{f(x+1)}{f(1)} \right)=\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

  24. anonymous
    • 5 years ago
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    Then\[\lim_{x->\infty}\log y=\lim_{x->\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

  25. anonymous
    • 5 years ago
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    Indeterminate, 0/0 (change that limit to 0, not infinity).

  26. anonymous
    • 5 years ago
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    Use L'Hopital's rule. Then\[\lim_{x \rightarrow 0}\frac{\log \frac{f(x+1)}{f(1)}}{x}=\frac{\lim_{x \rightarrow 0}\log \frac{f(x+1)}{f(1)}}{\lim_{x \rightarrow 0}x}\]\[=\frac{\lim_{x \rightarrow 0}\frac{\left( \frac{f'(x+1)}{f(1)} \right)}{\left( \frac{f(x+1)}{f(1)} \right)}}{\lim_{x \rightarrow 0}1} =\frac{\lim_{x \rightarrow 0}f'(x+1)}{\lim_{x \rightarrow 0}f(x+1)}\]

  27. anonymous
    • 5 years ago
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    No need to reply to this message. Just informing that I have received the notes

  28. anonymous
    • 5 years ago
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    \[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]

  29. anonymous
    • 5 years ago
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    I have got that. I can handle the rest.

  30. anonymous
    • 5 years ago
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    That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]

  31. anonymous
    • 5 years ago
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    Okay

  32. anonymous
    • 5 years ago
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    You are really a good person. In this world no one does so much for any other without any return

  33. anonymous
    • 5 years ago
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    Sometimes they do.

  34. anonymous
    • 5 years ago
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    This site has a good summary of limit laws: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

  35. anonymous
    • 5 years ago
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    Ok let me study the whole thing, I will let you know when I need any help. Thank you sooooooooo much

  36. anonymous
    • 5 years ago
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    Is that your university?

  37. anonymous
    • 5 years ago
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    No.

  38. anonymous
    • 5 years ago
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    Anyhow thanks

  39. anonymous
    • 5 years ago
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    np. good luck

  40. anonymous
    • 5 years ago
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    I don't want to be nosy about you, as it may offend you

  41. anonymous
    • 5 years ago
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    After all I am receiving something GREAT from you

  42. anonymous
    • 5 years ago
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    It's okay. Just do well in STEP.

  43. anonymous
    • 5 years ago
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    It will be a gift from me to you

  44. anonymous
    • 5 years ago
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    I will do good, and thats a promise

  45. anonymous
    • 5 years ago
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    Well, it's worth it then :)

  46. anonymous
    • 5 years ago
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    Ok bye now

  47. anonymous
    • 5 years ago
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    See you.

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