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Math Help - Can someone explain why this is?

  1. #1
    Junior Member ImaCowOK's Avatar
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    Can someone explain why this is?

    Can someone explain why the limit of the function 1/sin²x is ∞ as x approaches 0(from both left and right).

    I'm just starting to learn this, so as simple an explanation as you can manage would be appreciated.

    What does the graph of 1/sin² x look like?
    Last edited by ImaCowOK; September 5th 2010 at 06:53 AM.
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  2. #2
    MHF Contributor
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    Since \displaystyle \lim_{x \to 0} \sin^2 x = 0. This means we can make \sin^2 x as close to zero as we wish by making x sufficiently close to 0. Thus we can make \dfrac{1}{\sin^2x} as large as we wish by making x sufficiently close to 0.

    We can be a little more precise with a \delta-\epsilon definition if you wish.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by ImaCowOK View Post
    ...What does the graph of 1/sin² x look like?
    you can download this free graphing software form this site ...

    Graph

    ... which can produce the graph of \displaystyle y = \frac{1}{\sin^2{x}} = csc^2{x}
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  4. #4
    Junior Member ImaCowOK's Avatar
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    Thanks, it makes a lot more sense now.
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  5. #5
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    Note that for x= 0.001, sin(x)= 0.00001745329, approximately, so that [tex]sin^2(x)= 0.000000000304617419755778[tex] and then \frac{1}{sin^2(x)}= 3282806350.

    Of course, if x= -0.001, sin(x)= -0.0000174329 and so sin^2(x) and \frac{1}{sin^2(x)} are exactly the same as above.
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  6. #6
    Junior Member ImaCowOK's Avatar
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    Yes, I already understand that a negative number squared becomes positive.
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