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Math Help - Proving a Function is Continuous

  1. #1
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    Proving a Function is Continuous

    Hi there,

    The following is a suggested exercise I'm doing for my Analysis class.

    Prove f:x \rightarrow x^2

    Here's what I have, but it's straight from my class notes. I'm not sure what to do next.

    f is continuous at a if:
    <br />
\forall \epsilon > 0 \quad \exists \delta > 0 \medspace s.t.<br />
    <br />
\forall x \in A, \quad ||x-a||<\delta \implies ||f(x) - f(a)||< \epsilon<br />

    Any ideas?
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  2. #2
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    Quote Originally Posted by manygrams View Post
    Hi there,

    The following is a suggested exercise I'm doing for my Analysis class.

    Prove f:x \rightarrow x^2

    Here's what I have, but it's straight from my class notes. I'm not sure what to do next.

    f is continuous at a if:
    <br />
\forall \epsilon > 0 \quad \exists \delta > 0 \medspace s.t.<br />
    <br />
\forall x \in A, \quad ||x-a||<\delta \implies ||f(x) - f(a)||< \epsilon<br />

    Any ideas?
    Notice that

    |f(x)-f(a)|=|x^2-a^2|=|x-a||x+a|\le |x+a|\delta

    How can we control the factor |x+a| ? can you think of a good upperbound?
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  3. #3
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    What do you mean by control? And where did the less than or equal to |x+a|delta come from? Sorry, I'm really having trouble with all of this. Switched from Social Sciences into Mathematics and Economics... the questions are a lot different than what I'm used to.
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  4. #4
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    I'm guessing |x-a| must be less than or equal to delta?
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  5. #5
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    You shouldn't be "guessing" that, it is the definition of "limit"- besides it makes no sense until you have said what delta is! You want to have |x^2- a^2|= |x+a||x- a|< \epsilon and you want to be able to guarentee that by choosing some \delta so that "if |x- a|< \delta then [tex]|x^2- a^2|< \epsilon[tex]".

    As TheEmptySet told you, the crucial part is that |x+a|. You can always rewrite |x-a||x+a|< \epsilon as |x-a|< \frac{\epsilon}{|x+a|} and the only reason why you could not choose \delta to be \frac{\epsilon}{|x+a|} is that |x+a| is a function of x, not a constant.

    So can you find some number, A, such that \frac{\epsilon}{A}< \frac{\epsilon}{|x+a|}? If so, then choosing \delta= \frac{\epsilon}{A} will work:
    If |x- a|< \delta= \frac{\epsilon}{A}< \frac{\epsilon}{|x+a|}, then, working backwards, we would have |x- a||x+ a|= |x^2- a^2|< \epsilon.

    And how do we find that A? We want \frac{\epsilon}{A}< \frac{\epsilon}{|x+ a|}. Assuming, for the moment, that A is positive, that is the same as A> |x+ a|. Well, of course, there is no A for which that is true for all x but we are talking about x being close to a since this is a limit as x goes to a. Lets require that the distance from x to a be no greater than a/2: |x- a|< a/2. (I am assuming that a> 0 here. I will leave the cases a< 0 and a= 0 to you.) That is the same as -a/2< x- a< a/2 so, adding 2a to each part, 2a- a/2= 3a/2< x+ a< 5a/2. If we choose A= 5a/2, that will be true and so we will have |x- a|< \frac{\epsilon}{A}= \frac{2\epsilon}{5a}.

    Normally, we are interested in the case that \epsilon is very small number but there is no requirement that it be small. If we were just to choose \delta= \frac{2\epsilon}{5a} then our requirement that |x-a|< a/2 might not be true. So what we do is choose \delta to be the smaller of \frac{2\epsilon}{5a} and \frac{a}{2}.

    Again, this is for a> 0. You do the cases a< 0 and a= 0.
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