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Math Help - Limits

  1. #1
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    Limits

    find this
    the picture
    Attached Thumbnails Attached Thumbnails Limits-untitled.gif  
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  2. #2
    Ted
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    Quote Originally Posted by rqeeb View Post
    find this


    the picture
    evaluate it along the path y=mx^3.
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  3. #3
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    Quote Originally Posted by Ted View Post
    evaluate it along the path y=mx^3.
    it will be (m^3)/(1+m^4)

    but if we use y=x it will be zero

    which one is the correct
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  4. #4
    Ted
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    Quote Originally Posted by rqeeb View Post
    it will be (m^3)/(1+m^4)

    but if we use y=x it will be zero

    which one is the correct
    No need to use y=x; since the value of the limit along the path y=mx^3 depends on m, then the limit has different values for different values of m.
    Hence, the limit does not exist.
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  5. #5
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    Quote Originally Posted by Ted View Post
    No need to use y=x; since the value of the limit along the path y=mx^3 depends on m, then the limit has different values for different values of m.
    Hence, the limit does not exist.
    thanks

    but if we use polar coordinates it will be zero.... why
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  6. #6
    Ted
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    Quote Originally Posted by rqeeb View Post
    thanks

    but if we use polar coordinates it will be zero.... why
    If you use polar coordinates, you will face:

    \lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12  }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{r^2sin^3(\theta)cos^3(\theta)}{r^8cos^{12}(\  theta)+sin^4(\theta)}

    What did you do for it ?
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  7. #7
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    Quote Originally Posted by Ted View Post
    If you use polar coordinates, you will face:

    \lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12  }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{r^2sin^3(\theta)cos^3(\theta)}{r^8cos^{12}(\  theta)+sin^4(\theta)}

    What did you do for it ?
    this

    \lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12  }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{0}{0+sin^4(\theta)}=0
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  8. #8
    Ted
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    Quote Originally Posted by rqeeb View Post
    this

    \lim_{r\to 0} \frac{r^6sin^3(\theta)cos^3(\theta)}{r^{12}cos^{12  }(\theta)+r^4sin^4(\theta)}=\lim_{r\to 0} \frac{0}{0+sin^4(\theta)}=0
    Nope. The problem here is that sin^4(\theta) can be zero.
    and this will make the limit =\frac{0}{0} which is indeterminate.
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  9. #9
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    Quote Originally Posted by Ted View Post
    Nope. The problem here is that sin^4(\theta) can be zero.
    and this will make the limit =\frac{0}{0} which is indeterminate.
    Is that right!!
    because our teacher told us that we deal with any term has theta as a constant.
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  10. #10
    Ted
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    Quote Originally Posted by rqeeb View Post
    Is that right!!
    because our teacher told us that we deal with any term has theta as a constant.
    and the constant could be zero.
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  11. #11
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    It could be
    Thanks 4 everything
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