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    Hello,

    How do I prove that x^2 is continuous at x=1 by using the epsilon-delta method?

    Working so far is:
    |x^2-1| = x+1|x-1|
    (since we'll be close to x=1 so x+1 > 0)
    and I don't know where to go from there.

    If someone could give me a proof I'd appreciate it. I have no examples so seeing one would help. Thanks.
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    (1+h)^2 - 1 = 2h + h^2.
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    (Original post by Usar)
    Hello,

    How do I prove that x^2 is continuous at x=1 by using the epsilon-delta method?

    Working so far is:
    |x^2-1| = x+1|x-1|
    (since we'll be close to x=1 so x+1 > 0)
    and I don't know where to go from there.

    If someone could give me a proof I'd appreciate it. I have no examples so seeing one would help. Thanks.
    I think you forgot the brackets around the x+1. Should be |x^2-1| = |x+1||x-1| Now you need to find a delta such that |x-1| < delta implies |x^2-1|< epsilon
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    Ok, is this right: delta = min{epsilon/3, 1}?

    Working is:
    |(x+1)(x-1)| &lt; K|x-1| for some K.

    To get K:
    Assume |x-1| &lt; 1, then 1 &lt; x+1 &lt; 3, so K = 3.

    Therefore |x-1| &lt; epsilon/3.
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    Value for delta is right, but what you've got isn't really a proof that x^2 is cts. You've never even said what relevant "epsilon" has to what you're doing.

    Ideally, a proof should start something like: "Take \epsilon &gt; 0. Pick \delta = \min(\epsilon /3 , 1). Then ..." and conclude "... and so |x^2 - 1| &lt; \epsilon".
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    Thanks. I shouldn't have said proof, just the method of getting a value for delta.
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    OK another question:
    if f(x) = x^2 for x \in [0,2] and f(x) = 0 for x \in (2,\infty), how can I prove that it's discontinuous at x=2 using the epsilon-delta definition and proof by contradiction? I can't write |f(x) - f(2)| and substitute something in for f(x) cause f(x) is different either side of x=2.

    Am I right in saying that I consider two cases: when x>2 and when x < 2 and find that they somehow lead to contradictions? (I've done this but am not sure if this is a valid method).
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    You only need to consider 1 case.


    Big Spoiler
    For any \delta &gt; 0, what is a lower bound for |f(2) - f(2+\delta / 2)|?
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    Oh so it leads to epsilon > (say) 3. I'm not sure how to spot "tricks" like that -- hope it comes with more familiarity. Thanks for the help.
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    Btw, why did you use delta/2 in that expression (as opposed to delta)?
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    Just in case you got confused because the normal formulation involves considering |f(x)-f(2)| for |x-2| &lt; \delta, and |2+\delta - 2| isn't actually less than delta.
 
 
 
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