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nickymarden

  • 2 years ago

A question from my test today...

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

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

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

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

  6. nickymarden
    • 2 years ago
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    yup. :)

  7. eigenschmeigen
    • 2 years ago
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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

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

  9. eigenschmeigen
    • 2 years ago
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    hmm

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

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

  12. eigenschmeigen
    • 2 years ago
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    aw ok

  13. eigenschmeigen
    • 2 years ago
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    @TuringTest

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

  15. nickymarden
    • 2 years ago
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    haha

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

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

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

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

  20. experimentX
    • 2 years ago
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    looks like 2

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

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

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

  24. eigenschmeigen
    • 2 years ago
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    derp

  25. eigenschmeigen
    • 2 years ago
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    sorry guys

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

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

  28. SmoothMath
    • 2 years ago
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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.

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

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

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

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

  33. nickymarden
    • 2 years ago
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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 ;/

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

  35. nickymarden
    • 2 years ago
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    i knoow.

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

  37. TuringTest
    • 2 years ago
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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

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

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

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

  41. SmoothMath
    • 2 years ago
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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.

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

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

  44. TuringTest
    • 2 years ago
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    agreed :)

  45. SmoothMath
    • 2 years ago
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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.

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

  47. nickymarden
    • 2 years ago
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    omg

  48. experimentX
    • 2 years ago
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    LOL..

  49. nickymarden
    • 2 years ago
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    thanks :)

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

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