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    does it make sense, what I did in the pic? I considered the Max value of sinx-sinx0.
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    (Original post by cooldudeman)
    does it make sense, what I did in the pic? I considered the Max value of sinx-sinx0.
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    It is unnecessary to use this delta=min(epsilon,2). Can you see why taking delta=epsilon will work? The "work" in this question is showing that |sin(x)-sin(x_0)|<=|x-x_0|, where did you show this?

    Also, the final delta<epsilon in your answer should be a \leq
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    (Original post by Shillington)
    It is unnecessary to use this delta=min(epsilon,2). Can you see why taking delta=epsilon will work? The "work" in this question is showing that |sin(x)-sin(x_0)|<=|x-x_0|, where did you show this?
    am I meant to say epsilon is greater than 2?
    |sin(x)-sin(x_0)|<=2
    2< |x-x_0|<epsilon but I get the feeling that this doesn't make sense


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    (Original post by cooldudeman)
    am I meant to say epsilon is greater than 2?
    |sin(x)-sin(x_0)|<=2
    2< |x-x_0|<epsilon but I get the feeling that this doesn't make sense


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    I don't quite get what you're asking. You don't say anything about epislon, it is given to you. You must then show that when delta is appropriately chosen, the resulting sequence of inequalities holds i.e.

    |x-x_0|<delta
    =>
    |sin(x)-sin(x_0)|<epsilon
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    (Original post by Shillington)
    I don't quite get what you're asking. You don't say anything about epislon, it is given to you. You must then show that when delta is appropriately chosen, the resulting sequence of inequalities holds i.e.

    |x-x_0|<delta
    =>
    |sin(x)-sin(x_0)|<epsilon
    this is all I can think of to show
    |sin(x)-sin(x_0)|<=|x-x_0|?
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    (Original post by cooldudeman)
    this is all I can think of to show
    |sin(x)-sin(x_0)|<=|x-x_0|?
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    In all likelihood it was a given in the question, have you got the question to hand?

    I can't think of a neat way of showing it otherwise, have you even come across the definition of sin(x)?
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    (Original post by Shillington)
    In all likelihood it was a given in the question, have you got the question to hand?

    I can't think of a neat way of showing it otherwise, have you even come across the definition of sin(x)?
    this is the original q. 1st

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    (Original post by cooldudeman)
    this is the original q. 1st

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    See here for methods of showing |sin(x)-sin(y)|<=|x-y|
    http://math.stackexchange.com/questi...us-on-mathbb-r
 
 
 
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