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Math Help - Calculus Inequality Conundrum

  1. #1
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    Calculus Inequality Conundrum

    We're learning about the episilon-delta definition of limits.

    Anyway a related question was given by the course. Given 0<|x-1|<2 what range of values can |2x-3| be?

    Well I've got this to 0<|2x-2|<4 but i really don't know what to do from here, and how to solidly demonstrate the range of values |2x-3| can be.

    I managed to show |2x-3|<5 in a very long brute force way but I feel like there has to be a better way to do it. Any ideas?
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  2. #2
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    Re: Calculus Inequality Conundrum

    Hey jeremy5561.

    Hint: Show that if 0 < |x| < b then 0 <|x-a| < b+a and if 0 < |x| < c then 0 < |dx| < cd.

    To get you started consider that 0 < x < b and -b < x < 0 for x > 0 and x < 0 respectively. If I add a this means that:

    0 < |x-a| < c which implies 0 < x-a < c and -c < a-x < 0. By shifting the a term out to the boundaries we get:

    0 < x < a+c and -c-a < -x < 0. This proves that we have 0 < |x-a| < c+a for a > 0.

    Try the same thing for the multiplication case and then combine those two results to show a proof.
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    Re: Calculus Inequality Conundrum

    Quote Originally Posted by jeremy5561 View Post
    We're learning about the episilon-delta definition of limits.

    Anyway a related question was given by the course. Given 0<|x-1|<2 what range of values can |2x-3| be?

    Well I've got this to 0<|2x-2|<4 but i really don't know what to do from here, and how to solidly demonstrate the range of values |2x-3| can be.

    I managed to show |2x-3|<5 in a very long brute force way but I feel like there has to be a better way to do it. Any ideas?
    The triangle inequality is useful here... \displaystyle \begin{align*} |a + b| \leq |a| + |b| \end{align*}. So

    \displaystyle \begin{align*} |2x - 3| &\leq |2x-2| + |-1| \\ &\leq |2x - 2| + 1 \\ &\leq 4 + 1 \textrm{ since } |2x - 2| < 4 \\ &\leq 5 \end{align*}
    Thanks from jeremy5561
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