Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

GChamon

  • 2 years ago

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

  • This Question is Closed
  1. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    rewrite it as: e^(ln(1 + 1/2x)x) then manipulate it into l'hopital

  3. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  4. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I'll quickly get a pen and try

  5. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Just making sure, it's 2x in the denominator, right?

  6. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah is 1 over 2*x. this is why it is bugging me cuz i can't get rid of the indetermination

  7. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it \[\lim_{x\to \infty}(1+\frac{1}{2x})^x\]?

  8. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

  9. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you mean the limit to zero?

  10. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok think of it as \[(1+\frac{1}{2}\times \frac{1}{x})^x\]

  11. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    as x tends to 0

  12. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops ok lets try again

  13. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and no l'hopital??

  14. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    lol apparently not hey...

  15. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is cruel

  16. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ikr... such a wonderful tool. is like deriving using the concept of rate of change

  17. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok maybe replace \(x\) by \(\frac{1}{z}\)

  18. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then take the limit as z goes to infinity

  19. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, so using l'hôspital, is the answer 1?

  20. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I think I got it. Will explain now

  21. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh lord i am in idiot!

  22. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but then we get\[\exp\left(\lim_{z\to \infty}{\ln(1+\frac z2)\over z}\right)\]

  23. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you're an idiot satellite I hate to think what that makes me !

  24. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[e^{x\ln(1+\frac{1}{2x}}=e^{x\ln(\frac{2x+1}{2x})}\]

  25. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    tak the limit as x goes to zero oh wait nope not so snappy, ok i am not such an idiot

  26. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    That will only result in another indeterminate form

  27. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right i see that. for a moment i thought it was \(\ln(1)=0\) but it is not damn!

  28. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    kk I got it, will post method now:

  29. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    go slaiibak!

  30. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can't wait! no derivatives right?

  31. Kreshnik
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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....

  32. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol that is what i got the first try, but i thought it was as x goes to infinity too !

  33. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    It does have a derivative in, but it's not L"Hospital. Or it's arguable, but will post quickly..

  34. Kreshnik
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73 LOL

  35. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good lord... i'll have to spend these two days studying my retriceoff. have test on wednesday

  36. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whould it help to work the problem backwards? (from the answer)

  37. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I doubt it

  38. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[\exp(xln(2x+1)) = \exp({x^2 \ln(2x + 1) \over x})\] 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 \[\exp( \lim_{x \rightarrow 0} {x^2\ln(2x + 1) - (0)^2 \ln(2(0) + 1)\over x - 0} ) = \exp(f'(0))\]

  39. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we can all get the answer with l'hospital, but you said \(no\) \(l'hospital\) so we are doing loops for you

  40. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    f'(x) = 2x(ln(2x+1) + x^2 / (2x + 1) f'(0) = 0 therefore e^0 = 1. So the limit = 1

  41. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, that's a proof all right I guess since it doesn't technically use l'Hospital it's okay nice one!

  42. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  43. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  44. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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!!

  45. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow that was pretty snazzy!

  46. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha, got my english from a brit school (I am brazillian)

  47. slaaibak
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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!

  48. GChamon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks! that cheers me little to study! they all told me engineering is not easy. starting to see they were all right

  49. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.