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Thread: One-sided limit

  1. #1
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    One-sided limit

    Hi, geniuses!

    I need your expertise once again now that I've encountered a problem that I can't really figure out and Google wasn't doing much to help.

    I have to figure out what the limit of 1/x-cos(x)/sin(x) when x approaches pi-.

    I know that the answer is infinity but when i try to calculate it i always get negative infinity.
    My calculations are that 1/pi is a meaningless constant leaving us with -cos(x)/sin(x). cos(pi)=-1 and sin(pi)=0 so we have that - (-1) divided by an extremely small (negative, as it comes from -pi) number is equal to infinity. How is that possible? Are there any other ways to argue for the limit is infinity?

    Thanks a lot! Any answer will be appreciated!
    //DanishKid

  2. #2
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    You're oh so close.

    sin(pi-) > 0
    sin(pi+) < 0

    Chew on that.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
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    Quote Originally Posted by tkhunny View Post
    You're oh so close.

    sin(pi-) > 0
    sin(pi+) < 0

    Chew on that.
    Maybe I'm misunderstanding this, but I look at the unit circle and see that if I go to pi+ (counterclockwise) then sin has a positive number as its slightly over the x-axis, but if I go to pi- I'll have to go clockwise around the circle and therefore end up in a negative number. Am I wrong on this?

  4. #4
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    Quote Originally Posted by danishkid View Post
    Hi, geniuses!

    I need your expertise once again now that I've encountered a problem that I can't really figure out and Google wasn't doing much to help.

    I have to figure out what the limit of 1/x-cos(x)/sin(x) when x approaches pi-.

    I know that the answer is infinity but when i try to calculate it i always get negative infinity.
    My calculations are that 1/pi is a meaningless constant leaving us with -cos(x)/sin(x). cos(pi)=-1 and sin(pi)=0 so we have that - (-1) divided by an extremely small (negative, as it comes from -pi) number is equal to infinity. How is that possible? Are there any other ways to argue for the limit is infinity?

    Thanks a lot! Any answer will be appreciated!
    //DanishKid
    Is your problem:

    [tex]\displaystyle{\lim_{x \to {\pi}^{-}}\left [\frac{1}{x} - \frac{cos(x)}{sin(x)}\right ]}[/tex]...................this is what you posted

    or something else?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
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    Quote Originally Posted by Subhotosh Khan View Post
    Is your problem:

    [tex]\displaystyle{\lim_{x \to {\pi}^{-}}\left [\frac{1}{x} - \frac{cos(x)}{sin(x)}\right ]}[/tex]...................this is what you posted

    or something else?
    Yes! That's my problem

  6. #6
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    "pi-" has values less than pi. Approaching from an anti-clockwise direction if you must. Approaching from the left on a number line with typical orientation.
    "pi+" has values greater than pi. Approaching from a clockwise direction if you must. Approaching from the right on a number line with typical orientation.

    sin(pi-) > 0

    If you look at a number line, rather than a circle, it will be more clear. The circle definition doesn't mean much for this problem, anyway. Leave it and move to the number line.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  7. #7
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    Quote Originally Posted by tkhunny View Post
    "pi-" has values less than pi. Approaching from an anti-clockwise direction if you must. Approaching from the left on a number line with typical orientation.
    "pi+" has values greater than pi. Approaching from a clockwise direction if you must. Approaching from the right on a number line with typical orientation.

    sin(pi-) > 0

    If you look at a number line, rather than a circle, it will be more clear. The circle definition doesn't mean much for this problem, anyway. Leave it and move to the number line.
    I think I got it! Thank you very much!

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