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DLS

  • 2 years ago

Limits question

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  1. DLS
    • 2 years ago
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    \[\LARGE \lim_{x \rightarrow 0} (\frac{e^x+e^{-x}-2}{x^2})^\frac{1}{x^2}\]

  2. DLS
    • 2 years ago
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    @Callisto @satellite73

  3. DLS
    • 2 years ago
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    Summon @ParthKohli

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

  5. DLS
    • 2 years ago
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    I THOUGHT U CRACKED IT >.<

  6. Callisto
    • 2 years ago
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    Questions from you guys are nightmare for me!

  7. DLS
    • 2 years ago
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    Summon @terenzreignz

  8. terenzreignz
    • 2 years ago
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    Agree with @Callisto but let's see how this plays out :)

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

  10. DLS
    • 2 years ago
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    the power gives 1/12 hence e^(1/12) is the answer.

  11. terenzreignz
    • 2 years ago
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    Isn't this pretty straightforward as far as L'Hôpital stuff is concerned?

  12. DLS
    • 2 years ago
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    \[\large { (\frac{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+-2-x^2}{x^2})\frac{1}{x^2}}\] =112

  13. DLS
    • 2 years ago
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    i mean 1/12*

  14. DLS
    • 2 years ago
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    sorry i don't know why i posted such a simple ques :P maybe m sleepy .-.

  15. DLS
    • 2 years ago
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    just stuck me now

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

  17. DLS
    • 2 years ago
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    I said simple because of my own conscience :P

  18. terenzreignz
    • 2 years ago
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    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... \[\Large \ln(L) = \lim_{x\rightarrow 0} \ \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)^{\frac1{x^2}}\] \[\Large \ln(L) = \lim_{x\rightarrow 0}\frac1{x^2}\cdot \ln\left(\frac{e^x +e^{-x}-2}{x^2}\right)\]

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

  20. DLS
    • 2 years ago
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    LH twice? D:

  21. DLS
    • 2 years ago
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    hmm possibleway

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

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

  24. DLS
    • 2 years ago
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    lol

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

  26. DLS
    • 2 years ago
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    lengthy \\m// my method was better! :O

  27. terenzreignz
    • 2 years ago
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    What was your method? I didn't catch it :D

  28. DLS
    • 2 years ago
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    expansions..

  29. terenzreignz
    • 2 years ago
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    Series?

  30. DLS
    • 2 years ago
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    yes

  31. terenzreignz
    • 2 years ago
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    Count me out of that one :P

  32. DLS
    • 2 years ago
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    hmm :P it was a single line solution with expansions,LH is deadly dangerous here.

  33. terenzreignz
    • 2 years ago
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    That is true... unfortunately, I'm not well-versed with series... I don't like series, series don't like me :D

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