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    Can someone explain to me why we dont have an asymptote for x=0. Since the function is undefined. But how come it crosses at (0,1)? Like (cosx)/x has an asymptote there how come sine doesnt?

    Is it something to do with x tending towards infinity?
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    If you look at the Maclaurin series for sin(x) you will see, why sinx/x=1 at x=0 and for cosine, look at its Maclaurin series.
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    x = 0
    0 = number
    number has 6 letters
    6 + 0 = 6
    6,6,6 = 666

    satan

    ???
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    In case you cba http://mathworld.wolfram.com/MaclaurinSeries.html
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    It's called a "removable singularity", which is a term that you might want to look up
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    Whats that supposed to show? I thought maclaurin was just an expansion like the binomial, used to approximate things?
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    (Original post by Super199)
    Can someone explain to me why we dont have an asymptote for x=0. Since the function is undefined. But how come it crosses at (0,1)? Like (cosx)/x has an asymptote there how come sine doesnt?

    Is it something to do with x tending towards infinity?
    Something for you to think about,

    \displaystyle \int_{0}^{\infty} \frac{sin(x)}{x} dx
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    It is possible to prove that the limit as sinx/x approaches 0 is 1 by the squeeze (or sandwhich?) theorem.
    You can also express sinx as an infinite expansion (maclaurin series)
    sinx=x-x^3/3! + o(x^5)
    Then it follows that sinx/x = 1 - x^2/3! + o(x^4)
    This shows that as x approaches 0 sinx/x approaches 1, although it is important to note that x=0 is a discontinuity as this gives an undefined result.
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    (Original post by zetamcfc)
    Something for you to think about,

    \displaystyle \int_{0}^{\infty} \frac{sin(x)}{x} dx
    π/2
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    (Original post by Super199)
    Whats that supposed to show? I thought maclaurin was just an expansion like the binomial, used to approximate things?
    It's an approximation that approaches the actual function as the number of terms tends to infinity. If you take the expansion of sin(x) and divide it by x, you get 1 followed by lots of x terms. No matter how many terms you take, they will all still contain a power of x apart from the first term (1). Which means that when x=0, every single term containing x vanishes so you're just left with 1.
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    (Original post by B_9710)
    It is possible to prove that the limit as sinx/x approaches 0 is 1 by the squeeze (or sandwhich?) theorem.
    You can also express sinx as an infinite expansion (maclaurin series)
    sinx=x-x^3/3! + o(x^5)
    Then it follows that sinx/x = 1 - x^2/3! + o(x^4)
    This shows that as x approaches 0 sinx/x approaches 1, although it is important to note that x=0 is a discontinuity as this gives an undefined result.
    Ah that clears things up. Cheers
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    (Original post by Plagioclase)
    It's an approximation that approaches the actual function as the number of terms tends to infinity. If you take the expansion of sin(x) and divide it by x, you get 1 followed by lots of x terms. No matter how many terms you take, they will all still contain a power of x apart from the first term (1). Which means that when x=0, every single term containing x vanishes so you're just left with 1.
    Yh got it cheers
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    (Original post by B_9710)
    π/2
    Challenge, get there from the gamma integral :O
 
 
 
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