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AmTran_Bus

  • one year ago

Help finding crazy limit!

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  1. AmTran_Bus
    • one year ago
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  2. AmTran_Bus
    • one year ago
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    I need to find the limit as x approaches zero of (e^x -6)/(sin 4x) The problem is the online homework wants the limit as x approaches 0, but don t only the left and right hand limits exist? Check the atached image for the solution choices they provide. I thought it was negative infinity for lim as x approaches 0+ and infinity for x approaches 0-.

  3. ganeshie8
    • one year ago
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    You're correct. Looks there is a typographical error, the exponent 4 should be in the exponent i guess http://www.wolframalpha.com/input/?i=lim%28x%5Cto+0%29%5Cfrac%7B%5Cleft%28e%5Ex-6%5Cright%29%7D%7B%5Csin%5E4+%5Cleft%28x%5Cright%29%7D

  4. Michele_Laino
    • one year ago
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    hint: we can write this: \[\Large \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \frac{{{e^x} - 6}}{{4x}}\frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}}\]

  5. AmTran_Bus
    • one year ago
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    Oh ok. Thanks to both of you.

  6. AmTran_Bus
    • one year ago
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    Wait, isnt that going to come out as one @Michele_Laino And ganeshie8 said -inf?

  7. AmTran_Bus
    • one year ago
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    My mistake, I miscalculated @Michele_Laino

  8. anonymous
    • one year ago
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    Doing what Michele did helps you get rid of the sin(4x) portion of it. Which just makes it easier to calculate the left and right hand limits to show that it is \(-\infty\)

  9. ganeshie8
    • one year ago
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    it is -infy if you let sin(4x) be sin^4(x) but as the expression stands, the limit is DNE as you said

  10. AmTran_Bus
    • one year ago
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    ans was 1/4 :(

  11. AmTran_Bus
    • one year ago
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    Oh well, I can retake it.

  12. anonymous
    • one year ago
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    Oops, didnt recognize ganeshie's wolfram link was for sin^4(x), lol

  13. Michele_Laino
    • one year ago
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    I don't understand: 4 is an exponent or it is a factor?

  14. AmTran_Bus
    • one year ago
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    It was like the attachment said.

  15. Michele_Laino
    • one year ago
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    it is a factor, as I can see

  16. Michele_Laino
    • one year ago
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    in that case, your limit doesn't exist since we have to consider the graph of the fuinction 1/x, namely: |dw:1433615875097:dw|

  17. Michele_Laino
    • one year ago
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    I think we have this: \[\large \begin{gathered} \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = - \infty \hfill \\ \hfill \\ \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 - } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = + \infty \hfill \\ \end{gathered} \]

  18. anonymous
    • one year ago
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    Yeah, that is what you have. Something wrong with the question. You would think it was meant to be one of the variations ganeshie posted.

  19. Michele_Laino
    • one year ago
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    please substitute x=0.01, into your original expression, what do you get?

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