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GChamon Group Title

want some help with this Limit lim(x->0) (1+1/(2*x))^x at least a guidance! thanks!

  • 2 years ago
  • 2 years ago

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  1. GChamon Group Title
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    \[\lim_{x \rightarrow 0} (1+1/2x)^x\]

    • 2 years ago
  2. slaaibak Group Title
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    rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital

    • 2 years ago
  3. GChamon Group Title
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    thing is, we still haven't got to l'hôspital even though I know what ur talkign about. Teacher is not gonna consider any answer using l'hôspital... a real setback to these exercises. anyway to solve this without this tool?

    • 2 years ago
  4. slaaibak Group Title
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    I'll quickly get a pen and try

    • 2 years ago
  5. slaaibak Group Title
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    Just making sure, it's 2x in the denominator, right?

    • 2 years ago
  6. GChamon Group Title
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    yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination

    • 2 years ago
  7. satellite73 Group Title
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    is it \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?

    • 2 years ago
  8. GChamon Group Title
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    yeah

    • 2 years ago
  9. slaaibak Group Title
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    you mean the limit to zero?

    • 2 years ago
  10. satellite73 Group Title
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    ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]

    • 2 years ago
  11. GChamon Group Title
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    as x tends to 0

    • 2 years ago
  12. satellite73 Group Title
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    oops ok lets try again

    • 2 years ago
  13. satellite73 Group Title
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    and no l'hopital??

    • 2 years ago
  14. slaaibak Group Title
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    lol apparently not hey...

    • 2 years ago
  15. TuringTest Group Title
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    that is cruel

    • 2 years ago
  16. GChamon Group Title
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    ikr... such a wonderful tool. is like deriving using the concept of rate of change

    • 2 years ago
  17. satellite73 Group Title
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    ok maybe replace \(x\) by \(\frac{1}{z}\)

    • 2 years ago
  18. satellite73 Group Title
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    then take the limit as z goes to infinity

    • 2 years ago
  19. GChamon Group Title
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    ok, so using l'hôspital, is the answer 1?

    • 2 years ago
  20. slaaibak Group Title
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    I think I got it. Will explain now

    • 2 years ago
  21. satellite73 Group Title
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    oh lord i am in idiot!

    • 2 years ago
  22. TuringTest Group Title
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    but then we get\[\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)\]

    • 2 years ago
  23. TuringTest Group Title
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    if you're an idiot satellite I hate to think what that makes me !

    • 2 years ago
  24. satellite73 Group Title
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    \[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]

    • 2 years ago
  25. satellite73 Group Title
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    tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot

    • 2 years ago
  26. slaaibak Group Title
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    That will only result in another indeterminate form

    • 2 years ago
  27. satellite73 Group Title
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    right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!

    • 2 years ago
  28. slaaibak Group Title
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    kk I got it, will post method now:

    • 2 years ago
  29. TuringTest Group Title
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    go slaiibak!

    • 2 years ago
  30. satellite73 Group Title
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    can't wait! no derivatives right?

    • 2 years ago
  31. Kreshnik Group Title
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    You're supposed to know this rule: \[\LARGE \lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x=e\] \[\Large \lim_{x\to 0} \left(1+\frac{1}{2x}\right)^x\] \[\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{\frac{2x}{1}\cdot \frac{1}{2x }}\right]^x\] \[\Large \lim_{x\to 0} \left[\left(1+\frac{1}{2x}\right)^{2x} \right]^{\frac{x}{2x}}\] \[\LARGE \left(e\right)^{\frac12}=\sqrt e\] AAhhh.. LOOK WHAT I'VE DONE !! NOW I SEE THAT IT'S NOT INFINITE BUT IS "0" AUFFF... ( at least I'll post it :P ) hahahaah....

    • 2 years ago
  32. satellite73 Group Title
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    lol that is what i got the first try, but i thought it was as x goes to infinity too !

    • 2 years ago
  33. slaaibak Group Title
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    It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..

    • 2 years ago
  34. Kreshnik Group Title
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    @satellite73 LOL

    • 2 years ago
  35. GChamon Group Title
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    good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday

    • 2 years ago
  36. GChamon Group Title
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    whould it help to work the problem backwards? (from the answer)

    • 2 years ago
  37. TuringTest Group Title
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    I doubt it

    • 2 years ago
  38. slaaibak Group Title
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    \[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\] now using this: \[\lim_{x \rightarrow a} {f(x) - f(a) \over x - a} = f'(a)\] setting f(x) = x^2ln(2x + 1) we get \[\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1) - (0)^2 \ln(2(0) + 1)\over x - 0} ) = \exp(f'(0))\]

    • 2 years ago
  39. TuringTest Group Title
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    we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you

    • 2 years ago
  40. slaaibak Group Title
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    f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1

    • 2 years ago
  41. TuringTest Group Title
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    well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!

    • 2 years ago
  42. GChamon Group Title
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    would anyone bother to talk me a little bit through the proof? exp(xln(2x+1))=exp(x2ln(2x+1)x) how do u get to this in the first place? i know it sounds dumb but i am new to calculus

    • 2 years ago
  43. slaaibak Group Title
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    It's just a little bit of algebraic manipulation to force it into the f'(a) form. Check if you simplify, you will get back to the original thing.

    • 2 years ago
  44. GChamon Group Title
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    alright! gonna put some effort into learning more about functions since it is somehow not so clear to me how to manipulate the original expression. nevertheless, surely u guys helped a lot! thanks!!

    • 2 years ago
  45. satellite73 Group Title
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    wow that was pretty snazzy!

    • 2 years ago
  46. GChamon Group Title
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    haha, got my english from a brit school (I am brazillian)

    • 2 years ago
  47. slaaibak Group Title
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    GChamon, as you do more problems and exercises, you'll start to see it much clearer. Since you're only in the start of calculus now, you'll learn so much more and these things will become more intuitive. Good luck with the test!

    • 2 years ago
  48. GChamon Group Title
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    thanks! that cheers me little to study! they all told me engineering is not easy. starting to see they were all right

    • 2 years ago
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