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AmTran_Bus

  • one year ago

Limit question?

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  1. AmTran_Bus
    • one year ago
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    |dw:1436286205012:dw|

  2. AmTran_Bus
    • one year ago
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    Um. Here are my choices but Im getting DNE

  3. AmTran_Bus
    • one year ago
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    Just had an idea

  4. AmTran_Bus
    • one year ago
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    What in the world someone help! http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+%28e%5Ex-6%29%2Fsin+4x

  5. AmTran_Bus
    • one year ago
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    The problem does not say from the left or right!!!!!!!!! So my DNE is right but what do I do?

  6. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{e^x-6}{\sin(4x)}}\) ?

  7. AmTran_Bus
    • one year ago
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    Correct

  8. welshfella
    • one year ago
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    would l'hopitals rule be helpful here?

  9. AmTran_Bus
    • one year ago
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    Good thought. Let me try it

  10. SolomonZelman
    • one year ago
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    can't do it yet

  11. SolomonZelman
    • one year ago
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    it is not 0/0 when you plug in x=0

  12. welshfella
    • one year ago
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    right - only the denoiminator = 0 l'hopitals no help

  13. AmTran_Bus
    • one year ago
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    Oh yea, must be inf/inf or 0/0

  14. welshfella
    • one year ago
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    as x approaches 0 sin 4x approaches 0 and e^x - 6 approaches -5 so can we then say that the limit is - infinity?

  15. welshfella
    • one year ago
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    - approaching from above 0

  16. SolomonZelman
    • one year ago
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    you are right, the limit actually doesn't exst.

  17. AmTran_Bus
    • one year ago
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    Yea, thats what has thrown me off for a long time. But wolfram treats it as 0+ and 0- and and the answer choices have both neg inf and pos inf

  18. SolomonZelman
    • one year ago
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    even one sides limits diverge

  19. SolomonZelman
    • one year ago
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    when you get a limit that is equal to +∞ or -∞, that means the limit does NOT exist. (It is infinite - not limited)

  20. AmTran_Bus
    • one year ago
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    But it is an ans choice!

  21. SolomonZelman
    • one year ago
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    Well, for two sides limit, it DNE

  22. AmTran_Bus
    • one year ago
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    I agree

  23. SolomonZelman
    • one year ago
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    from the right side it is ∞, from the left side it is -∞

  24. AmTran_Bus
    • one year ago
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    dumb problem and ans choices

  25. SolomonZelman
    • one year ago
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    do you have an answer choice does not exist?

  26. AmTran_Bus
    • one year ago
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  27. welshfella
    • one year ago
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    yes near the y axis the graph looks like |dw:1436287092439:dw|

  28. AmTran_Bus
    • one year ago
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    @SolomonZelman

  29. AmTran_Bus
    • one year ago
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    @welshfella so do you agree DNE?

  30. welshfella
    • one year ago
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    yes

  31. AmTran_Bus
    • one year ago
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    Well thanks for all your help! I may email prof

  32. AmTran_Bus
    • one year ago
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    Well, I asked on yahoo ans and they said -inf. I just dont see it but gotta mark something @hartnn @SolomonZelman @welshfella

  33. hartnn
    • one year ago
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    limit shouldn't exist sin 4x is negative when x<0 and it is positive when x>0 hence Left hand limit would not = right hand limit

  34. AmTran_Bus
    • one year ago
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    I agree hartnn. Just these ans choices giving fit.

  35. SolomonZelman
    • one year ago
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    ok, maybe the question is a little bit different?

  36. AmTran_Bus
    • one year ago
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    ??

  37. SolomonZelman
    • one year ago
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    Are you sure it is exactly saying \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{e^x-6}{\sin(4x)}}\)

  38. SolomonZelman
    • one year ago
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    well, then this limit doesn't exist.

  39. SolomonZelman
    • one year ago
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    I mean if limit=∞ or -∞ it doesn't exist (either way)

  40. hartnn
    • one year ago
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    people saying -infinity are assuming that sin4x = 0 e^x -6 will be negative

  41. SolomonZelman
    • one year ago
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    ohh...

  42. SolomonZelman
    • one year ago
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    yeah, should have noticed that:)

  43. SolomonZelman
    • one year ago
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    Bus?

  44. AmTran_Bus
    • one year ago
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    I'm here! So is it -inf after all?

  45. SolomonZelman
    • one year ago
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    yes

  46. AmTran_Bus
    • one year ago
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    Sin 4x=0?

  47. SolomonZelman
    • one year ago
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    yes

  48. hartnn
    • one year ago
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    no lol, the limit is still DNE. In yahoo answer, they might have said -inf as they might have assumed sin4x =0 ... that was my point, but sin 4x is not actually 0

  49. SolomonZelman
    • one year ago
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    well, if you take the limit of sin(4x) as x→0 alone, then it is 0

  50. AmTran_Bus
    • one year ago
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    Story of my life...

  51. hartnn
    • one year ago
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    why would you do that lol...

  52. SolomonZelman
    • one year ago
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    1 D, XD = (1+x)D

  53. SolomonZelman
    • one year ago
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    jk

  54. SolomonZelman
    • one year ago
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    idk, hartnn

  55. AmTran_Bus
    • one year ago
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    Gonna email prof

  56. SolomonZelman
    • one year ago
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    the limit is indeterminate, beause you are trying to divide by zero

  57. SolomonZelman
    • one year ago
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    technically u r...

  58. SolomonZelman
    • one year ago
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    when you divide by a number that approaches 0 from the right, you get ∞, and when you divide by a number that approaches 0 from the left, you get -∞. if the divident is positive. In this case divident is negative, so it is the other way around. when you divide by a number that approaches 0 from the right, you get -∞, and when you divide by a number that approaches 0 from the left, you get ∞. And if you take a 2-sides limit (as in your case), you shouldn't get just ∞ or -∞.....

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