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schmidtdancer

  • 3 years ago

FIND the limit!!

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  1. schmidtdancer
    • 3 years ago
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    |dw:1343661876978:dw|

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

  3. schmidtdancer
    • 3 years ago
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    ok

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

  5. ParthKohli
    • 3 years ago
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    Let's see how the graph is..

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

  7. ParthKohli
    • 3 years ago
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    Don't do that now.

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

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

  10. schmidtdancer
    • 3 years ago
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    Ok

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

  12. agentx5
    • 3 years ago
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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)

  13. schmidtdancer
    • 3 years ago
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    ok! can you show how I would solve it

  14. schmidtdancer
    • 3 years ago
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    Ok i understand that

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

  16. schmidtdancer
    • 3 years ago
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    So how would I solve the limit?

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

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

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

  20. schmidtdancer
    • 3 years ago
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    2.99(9-2.99^2)

  21. ParthKohli
    • 3 years ago
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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.

  22. schmidtdancer
    • 3 years ago
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    for 2.5, f(x) would be 4.145

  23. ParthKohli
    • 3 years ago
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    Find them for all.

  24. agentx5
    • 3 years ago
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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.

  25. schmidtdancer
    • 3 years ago
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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

  26. ParthKohli
    • 3 years ago
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    Calculator, Schmidt.

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

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

  29. schmidtdancer
    • 3 years ago
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    I got 7.3484

  30. ParthKohli
    • 3 years ago
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    Just not 3.

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

  32. schmidtdancer
    • 3 years ago
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    Ok! I think its going to be 7.5

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

  34. schmidtdancer
    • 3 years ago
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    I just tried for that

  35. schmidtdancer
    • 3 years ago
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    can someone tell me if im right

  36. ParthKohli
    • 3 years ago
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    I am too lazy to do so :/

  37. schmidtdancer
    • 3 years ago
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    come onnn lol

  38. schmidtdancer
    • 3 years ago
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    plz! i really need your help..

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

  40. schmidtdancer
    • 3 years ago
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    OK

  41. ParthKohli
    • 3 years ago
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    @agentx5 Where is the graph?

  42. agentx5
    • 3 years ago
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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|

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

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

  45. ParthKohli
    • 3 years ago
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    The limit was getting smaller and smaller. ;)

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

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

  48. agentx5
    • 3 years ago
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    With the limit shown... |dw:1343663226621:dw|

  49. schmidtdancer
    • 3 years ago
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    like 2 or 3 right?

  50. agentx5
    • 3 years ago
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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. ;-)

  51. schmidtdancer
    • 3 years ago
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    Guys do you know the actual limit?

  52. agentx5
    • 3 years ago
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    Yes.

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

  54. schmidtdancer
    • 3 years ago
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    Yeah I do see it visually,

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

  56. schmidtdancer
    • 3 years ago
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    wait it looks around 0

  57. agentx5
    • 3 years ago
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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.

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

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

  60. agentx5
    • 3 years ago
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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.

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

  62. agentx5
    • 3 years ago
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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?

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

  64. agentx5
    • 3 years ago
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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. :-)

  65. schmidtdancer
    • 3 years ago
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    wait so the limit is DNE? or zero!?!

  66. agentx5
    • 3 years ago
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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.

  67. agentx5
    • 3 years ago
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    You will see DNE problems, I can guarantee you.

  68. schmidtdancer
    • 3 years ago
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    Oh yeah, I know!:)

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