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Math Help - Proving Continuity

  1. #1
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    Proving Continuity

    so im supposed to prove that x^2-7 is continuous at x=1

    f(1) = -6
    and now i need to use the formal definition to prove this.
    so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

    i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
    and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"
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    Quote Originally Posted by ah-bee View Post
    so im supposed to prove that x^2-7 is continuous at x=1

    f(1) = -6
    and now i need to use the formal definition to prove this.
    so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

    i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
    and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"
    There's an easy way of doing these. I'll have time to show it later.
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    awesome.. cant wait
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  4. #4
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    Quote Originally Posted by ah-bee View Post
    so im supposed to prove that x^2-7 is continuous at x=1

    f(1) = -6
    and now i need to use the formal definition to prove this.
    so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

    i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
    and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"
    1. |(x^2 - 7) - (-6)| = |x^2 - 1| = |x - 1| |x + 1|.


    2. Note that |x - 1| < \delta_0 \Rightarrow 1 - \delta_0 < x < 1 + \delta_0.
    Only values of x in the neighbourhood of x = 1 are of interest so let {\color{red}\delta_0 = \frac{1}{2}}.
    Then {\color{red}\frac{1}{2} < x < \frac{3}{2} \Rightarrow |x - 1| < \frac{1}{2}}.



    3. Under this restricted set of values of x, |x + 1| \leq \frac{5}{2} < 3.


    4. Therefore |x - 1| |x + 1| < 3 |x - 1| for |x - 1| < \frac{1}{2}.

    Note that |x - 1| |x + 1| < \epsilon \Rightarrow 3 |x - 1| < \epsilon \Rightarrow |x - 1| < \frac{\epsilon}{3}.


    5. Let \delta(\epsilon) = min \left \{ \frac{1}{2}, ~ \frac{\epsilon}{3}\right \}.

    Then for all \epsilon > 0 there exists \delta(\epsilon) > 0 such that 0 < |x - 1| < \delta(\epsilon) \Rightarrow |(x^2 - 7) - (-6)| < \epsilon.

    Therefore \lim_{x \rightarrow 1} (x^2 - 7) = -6.


    The key to freedom is steps 2-3 where you make a convenient restriction on the values of x to get an inequality involving only |x - 1| .....
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    thats great, thanks! but i was told by my lecturer to approach this type of question by supposing delta = something, and then deducing the |f(x)-f(a)|<epsilon. i dunno, might just be some technicalities but your method looks great to me, mainly because there isnt a delta that i can choose straight up and then do some algebraic manipulation to end up with |f(x)-f(a)|<epsilon.
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  6. #6
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    Quote Originally Posted by ah-bee View Post
    thats great, thanks! but i was told by my lecturer to approach this type of question by supposing delta = something, and then deducing the |f(x)-f(a)|<epsilon. i dunno, might just be some technicalities but your method looks great to me, mainly because there isnt a delta that i can choose straight up and then do some algebraic manipulation to end up with |f(x)-f(a)|<epsilon.
    If it survives 48 hours without major criticism from ThePerfectHacker, then you can take it to the bank .....
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