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nickymarden

A question from my test today...

  • 2 years ago
  • 2 years ago

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  1. nickymarden
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    • 2 years ago
  2. nickymarden
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    can't anyone do it? Where's @TuringTest

    • 2 years ago
  3. eigenschmeigen
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    can we factorise and cancel to get rid of the pesky denominator? :\[\frac{-x(x-1)(x+3)}{x-1} \le g(x) \le \frac{(x+1)(x-1)}{x-1}\]

    • 2 years ago
  4. nickymarden
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    i guess, i didnt have time to do it, Last question of the test... ;/

    • 2 years ago
  5. eigenschmeigen
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    so: \[-x(x+3) \le g(x) \le (x+1)\] and you want the limit as x tends to 1 ?

    • 2 years ago
  6. nickymarden
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    yup. :)

    • 2 years ago
  7. eigenschmeigen
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    well evaluating at x = 1 we have \[-4 \le g(x) \le 2\] im not sure whether thats what you want or not, i haven't really seen this kinda Q before

    • 2 years ago
  8. nickymarden
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    it isn't, there's only one value. ;/ But thank you :))

    • 2 years ago
  9. eigenschmeigen
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    hmm

    • 2 years ago
  10. eigenschmeigen
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    call in the cavalry! @FoolForMath

    • 2 years ago
  11. FoolForMath
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    My mind is officially fried for the day :( sorry.

    • 2 years ago
  12. eigenschmeigen
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    aw ok

    • 2 years ago
  13. eigenschmeigen
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    @TuringTest

    • 2 years ago
  14. TuringTest
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    whoa what happened? let em read this

    • 2 years ago
  15. nickymarden
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    haha

    • 2 years ago
  16. TuringTest
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    the limit as x goes to 1 ?

    • 2 years ago
  17. zepp
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    Apply the squeeze theorem to the question :)

    • 2 years ago
  18. TuringTest
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    what limit is this, I don't understand the notation

    • 2 years ago
  19. nickymarden
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    \[\lim_{x \rightarrow 1}\]

    • 2 years ago
  20. experimentX
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    looks like 2

    • 2 years ago
  21. TuringTest
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    I hope smoothmath's cookin up something good 'cuz I see what Eigen saw... how do you figure X ?

    • 2 years ago
  22. eigenschmeigen
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    i factorised wrong i think

    • 2 years ago
  23. TuringTest
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    oh I trusted you! ok I'll check

    • 2 years ago
  24. eigenschmeigen
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    derp

    • 2 years ago
  25. eigenschmeigen
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    sorry guys

    • 2 years ago
  26. experimentX
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    -(x-1)(x-3)

    • 2 years ago
  27. eigenschmeigen
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    yah. sorry im exhausted, just finished ~ 4hrs physics work

    • 2 years ago
  28. SmoothMath
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    I like l'hopitals rule for both of those equations. To use L'hopital's rule, we have to make sure they satisfy the condition that the functions at that point are of the indeterminate form 0/0 or inf/inf. These both look like 0/0, so we're good to go. Now, since we've satisfied that condition, L'hopital's rule says that the limit of each function will be the same as if we take the derivative of the top and bottom. So basically, if our function is f(x)/g(x), the rule says that the limit will be the same as f'(x)/g'(x), as long as that limit exists. What that gives us is: (-3x^2 + 8x - 3)/1 <= g(x) <= 2x/1 Evaluating at x=1 gives: (-3 + 8 -3)/1 <= g(x) <= 2/1 2 <=g(x) <=2 So, by squeeze theorem, lim g(x) is 2.

    • 2 years ago
  29. eigenschmeigen
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    in fact ima go sleep night!

    • 2 years ago
  30. SmoothMath
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    Haha don't worry, Turing. I cooked up something delicious.

    • 2 years ago
  31. TuringTest
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    I can see that! that's what I get for not double-checking other peoples algebra...

    • 2 years ago
  32. experimentX
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    well that was delicious.

    • 2 years ago
  33. nickymarden
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    good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/

    • 2 years ago
  34. SmoothMath
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    Ugh. Sucks. I hate that he disallowed it because it works so well...

    • 2 years ago
  35. nickymarden
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    i knoow.

    • 2 years ago
  36. experimentX
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    factorize and cancel ... that's quite easier that L'hospital's rule.

    • 2 years ago
  37. TuringTest
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    but it does factor to\[{-x(x-1)(x-3)\over x-1}\le g(x)\le x+1\]\[-x(x-3)\le g(x)\le x+1\]so yeah, we can just plug in like experimentX says

    • 2 years ago
  38. experimentX
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    well ... still, I really liked that squeezing theorem.

    • 2 years ago
  39. SmoothMath
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    Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.

    • 2 years ago
  40. experimentX
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    real mathematicians do that ..!!!

    • 2 years ago
  41. SmoothMath
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    But yeah, to solve by factoring, turing's method above is correct. Eigen's original misfactoring is what gave the error we got.

    • 2 years ago
  42. SmoothMath
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    Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.

    • 2 years ago
  43. experimentX
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    you are indeed a real one!!!

    • 2 years ago
  44. TuringTest
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    agreed :)

    • 2 years ago
  45. SmoothMath
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    Haha okay. Cool. Both methods are great, and I actually think factoring is easier to teach. I just find L'hopital's easier to execute. And sexier.

    • 2 years ago
  46. TuringTest
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    haha you should be "SexyMath" then

    • 2 years ago
  47. nickymarden
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    omg

    • 2 years ago
  48. experimentX
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    LOL..

    • 2 years ago
  49. nickymarden
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    thanks :)

    • 2 years ago
  50. SmoothMath
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    lols. That might give the wrong impression, but I like it.

    • 2 years ago
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