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schmidtdancer

FIND the limit!!

  • one year ago
  • one year ago

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  1. schmidtdancer
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    |dw:1343661876978:dw|

    • one year ago
  2. agentx5
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    Ah! This is one best though of visually, let me make you a graph:

    • one year ago
  3. schmidtdancer
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    ok

    • one year ago
  4. ParthKohli
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    If you find the function continuous, then you may just plug 3 for \(x\) by the way.

    • one year ago
  5. ParthKohli
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    Let's see how the graph is..

    • one year ago
  6. schmidtdancer
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    Ok. If I plug in 3, I get 3 sqrt(0)

    • one year ago
  7. ParthKohli
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    Don't do that now.

    • one year ago
  8. agentx5
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    What you need are numbers slightly less than 3, like 2.9999999999999

    • one year ago
  9. ParthKohli
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    Looks like we can plug in values less than \(3\).

    • one year ago
  10. schmidtdancer
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    Ok

    • one year ago
  11. ParthKohli
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    Because it is a left handed limit, we may plug numbers like agentx5 said :)

    • one year ago
  12. agentx5
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    That - sign above the limit means "numbers from the left hand side limit" or the number slightly less than 3 (but as close as you can get)

    • one year ago
  13. schmidtdancer
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    ok! can you show how I would solve it

    • one year ago
  14. schmidtdancer
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    Ok i understand that

    • one year ago
  15. telliott99
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    About your graph, notice what happens to the function for x > 3

    • one year ago
  16. schmidtdancer
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    So how would I solve the limit?

    • one year ago
  17. ParthKohli
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    By plugging numbers slightly less than 3.

    • one year ago
  18. agentx5
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    @telliott99 , I know... look at the notation for imaginary and real...

    • one year ago
  19. telliott99
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    Just saying the graph is not correct. It doesn't have a hole in it. And do what @ParthKohli says

    • one year ago
  20. schmidtdancer
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    2.99(9-2.99^2)

    • one year ago
  21. ParthKohli
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    \[ \begin{array}{l|r} x & f(x) \\ \hline 2.5 & \\ 2.9 \\ 2.999 \\ 2.99999\end{array} \] Fill the f(x)'s.

    • one year ago
  22. schmidtdancer
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    for 2.5, f(x) would be 4.145

    • one year ago
  23. ParthKohli
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    Find them for all.

    • one year ago
  24. agentx5
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    And and as you get closer and closer to 2.999999999999? Every closer to 3 from the "just under" side and you'll approach some limit.

    • one year ago
  25. schmidtdancer
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    Ok can you help find the limit now? Like I understand how to find that, i just dont haeve time to find them all

    • one year ago
  26. ParthKohli
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    Calculator, Schmidt.

    • one year ago
  27. schmidtdancer
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    so agentx5, I should just use 2.999999 to find the limit?

    • one year ago
  28. agentx5
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    Or if you prefer, 2.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999... Get the point? :P

    • one year ago
  29. schmidtdancer
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    I got 7.3484

    • one year ago
  30. ParthKohli
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    Just not 3.

    • one year ago
  31. ParthKohli
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    Looks like its getting closer to 7.5, but let's see it for more values.

    • one year ago
  32. schmidtdancer
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    Ok! I think its going to be 7.5

    • one year ago
  33. ParthKohli
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    Are you sure? Try for 2.9999999 and 2.999999999999999. So on.

    • one year ago
  34. schmidtdancer
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    I just tried for that

    • one year ago
  35. schmidtdancer
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    can someone tell me if im right

    • one year ago
  36. ParthKohli
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    I am too lazy to do so :/

    • one year ago
  37. schmidtdancer
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    come onnn lol

    • one year ago
  38. schmidtdancer
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    plz! i really need your help..

    • one year ago
  39. ParthKohli
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    All right :) let me check it out on Wolfram.

    • one year ago
  40. schmidtdancer
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    OK

    • one year ago
  41. ParthKohli
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    @agentx5 Where is the graph?

    • one year ago
  42. agentx5
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    Here's the algebra trick, you need to find what y would be when x=3. You can't do this normally but... \(x\sqrt{(3+x)(3-x)}\) , For what solutions does y = 0? Well, 0, +3 and -3. So... ;-) Replacing with a better graph (more accurate, real #'s only): |dw:1343662913297:dw|

    • one year ago
  43. schmidtdancer
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    ok thx @agentx5 so what would the limit be for this probb?

    • one year ago
  44. ParthKohli
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    I was actually surprised how I was plugging values and getting it approaching something else. Hence, I asked.

    • one year ago
  45. ParthKohli
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    The limit was getting smaller and smaller. ;)

    • one year ago
  46. ParthKohli
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    You may find from the graph, actually.|dw:1343663313029:dw|

    • one year ago
  47. ParthKohli
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    What is \(y\) approaching in the graph? ^

    • one year ago
  48. agentx5
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    With the limit shown... |dw:1343663226621:dw|

    • one year ago
  49. schmidtdancer
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    like 2 or 3 right?

    • one year ago
  50. agentx5
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    I actually had my imaginary lines going the wrong way @telliott99 and the circles were too high in the original sketch (they're on the x-axis), I stand corrected. Although I'm not sure that's what you meant. ;-)

    • one year ago
  51. schmidtdancer
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    Guys do you know the actual limit?

    • one year ago
  52. agentx5
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    Yes.

    • one year ago
  53. agentx5
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    And I'm trying to help you learn visually what's going on so you're not lost here

    • one year ago
  54. schmidtdancer
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    Yeah I do see it visually,

    • one year ago
  55. schmidtdancer
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    is the limit around 2 or 3

    • one year ago
  56. schmidtdancer
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    wait it looks around 0

    • one year ago
  57. agentx5
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    \[\lim_{x \rightarrow 3^-}\] What is the output of the function approaching? The input gets closer to 3 from the "slightly less" side.

    • one year ago
  58. schmidtdancer
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    as 3 approaches 3 from the left, the y value is 0, right?

    • one year ago
  59. schmidtdancer
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    sorry i meant, as x approaches 3

    • one year ago
  60. agentx5
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    Or using the analytical method, as you plug in larger and larger 2.9999999999999999...'s you get smaller and smaller numbers. Infinitesimally small. To the point gets damned close to zero. Thus the limit is zero, correct.

    • one year ago
  61. schmidtdancer
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    ok so the limit for this function is 0? :D

    • one year ago
  62. agentx5
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    The reason you can't just substitute 3 in place of x, officially is because you might have one like this: \[\lim_{x \rightarrow 1} = \frac{1}{1-x}\] Look what happens if you plug in 1?

    • one year ago
  63. schmidtdancer
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    But, limit is 0 right? and yes the ddenominator bneomes 0.. not possible

    • one year ago
  64. agentx5
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    In this case your left and your right hand limits are going in opposite directions, and the limit doesn't exist. |dw:1343663845294:dw| And yes, twice to the above post. :-)

    • one year ago
  65. schmidtdancer
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    wait so the limit is DNE? or zero!?!

    • one year ago
  66. agentx5
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    For your original problem, Zero. For the one I just asked you, trying to get you to learn, Does Not Exist.

    • one year ago
  67. agentx5
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    You will see DNE problems, I can guarantee you.

    • one year ago
  68. schmidtdancer
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    Oh yeah, I know!:)

    • one year ago
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