Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DLS Group Title

Limits question

  • one year ago
  • one year ago

  • This Question is Closed
  1. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\LARGE \lim_{x \rightarrow 0} (\frac{e^x+e^{-x}-2}{x^2})^\frac{1}{x^2}\]

    • one year ago
  2. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @Callisto @satellite73

    • one year ago
  3. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Summon @ParthKohli

    • one year ago
  4. ParthKohli Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    (Just that you don't get in hopes: I'm just trying out. @Callisto please feel free to post your answer!)

    • one year ago
  5. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I THOUGHT U CRACKED IT >.<

    • one year ago
  6. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Questions from you guys are nightmare for me!

    • one year ago
  7. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Summon @terenzreignz

    • one year ago
  8. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Agree with @Callisto but let's see how this plays out :)

    • one year ago
  9. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\LARGE e^{ (\frac{e^x+e^{-x}-2-x^2}{x^2})\frac{1}{x^2}}\] this should be it,I just need to find the limit of this thing,hmm..

    • one year ago
  10. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    the power gives 1/12 hence e^(1/12) is the answer.

    • one year ago
  11. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?

    • one year ago
  12. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large { (\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+-2-x^2}{x^2})\frac{1}{x^2}}\] =112

    • one year ago
  13. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    i mean 1/12*

    • one year ago
  14. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry i don't know why i posted such a simple ques :P maybe m sleepy .-.

    • one year ago
  15. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    just stuck me now

    • one year ago
  16. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I didn't say simple, I said straightforward :D \[\Large 1^\infty\] So... suppose... \[\Large L = \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\]

    • one year ago
  17. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I said simple because of my own conscience :P

    • one year ago
  18. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Then... \[\Large \ln(L) = \ln \lim_{x\rightarrow 0}\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\] Since Logarithms are continuous, they may enter the limit... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)\]

    • one year ago
  19. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\Large \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) =\lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}\] And now we have an indeterminate form \(\large\frac00\)

    • one year ago
  20. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    LH twice? D:

    • one year ago
  21. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm possibleway

    • one year ago
  22. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yup :) Differentiating the numerator and denominator... (bloody hell, this'll be a mouthful :D ) \[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

    • one year ago
  23. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\Large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

    • one year ago
  24. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    lol

    • one year ago
  25. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It keeps getting cropped >.> \[\large \lim_{x\rightarrow 0} \frac{\ln\left(\frac{e^x +e^{-x}-2}{x^2}\right) }{x^2}=\lim_{x\rightarrow0}\frac{\left(\frac{x^2}{e^x+e^{-x}-2}\right)\left(\frac{x^2(e^x-e^{-x})-2x(e^x+e^{-x}-2)}{x^4}\right)}{2x}\]

    • one year ago
  26. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    lengthy \\m// my method was better! :O

    • one year ago
  27. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    What was your method? I didn't catch it :D

    • one year ago
  28. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    expansions..

    • one year ago
  29. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Series?

    • one year ago
  30. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

    • one year ago
  31. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Count me out of that one :P

    • one year ago
  32. DLS Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm :P it was a single line solution with expansions,LH is deadly dangerous here.

    • one year ago
  33. terenzreignz Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    That is true... unfortunately, I'm not well-versed with series... I don't like series, series don't like me :D

    • one year ago
    • Attachments:

See more questions >>>

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.