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Math Help - Continuity proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Continuity proof

    Let g(x)=x^{1/3}

    Q:

    a) Prove that g is continuous at c=0

    b) Prove that g is continuous at a point c\neq\\0. (The identity a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) will be helpful.)

    A: a) Let \epsilon>0. Then, we have to show that, if |x-c|=|x-0|=|x|<\delta it follows that

    |g(x)-c|=|x^{1/3}-0|=|x^{1/3}|<\epsilon.

    So, if \delta=\epsilon, then |x|<\delta implies |x^{1/3}|<\epsilon. Therefore, g(x) is ontinuous at 0.

    b) Given I am not sure about the first part, I am pretty stuck on the (b). I am not sure how to use the identity in my scratch work; I keep hitting a brick wall. Some help would be great.

    Thanks
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    Let g(x)=x^{1/3}

    Q:

    a) Prove that g is continuous at c=0

    b) Prove that g is continuous at a point c\neq\\0. (The identity a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) will be helpful.)

    A: a) Let \epsilon>0. Then, we have to show that, if |x-c|=|x-0|=|x|<\delta it follows that

    |g(x)-c|=|x^{1/3}-0|=|x^{1/3}|<\epsilon.

    So, if \delta=\epsilon, then |x|<\delta implies |x^{1/3}|<\epsilon. Therefore, g(x) is ontinuous at 0.

    b) Given I am not sure about the first part, I am pretty stuck on the (b). I am not sure how to use the identity in my scratch work; I keep hitting a brick wall. Some help would be great.
    For (a), \delta=\varepsilon doesn't work. You need to take \delta=\varepsilon^3. That will ensure that \delta^{1/3}=\varepsilon.

    For (b), the idea is this. You want to show that if |x-c| is small enough then |x^3-c^3| will be small. But |x^{3}-c^{3}|=|x-c||x^{2}+xc+c^{2}|. So you have to concentrate on ensuring that x^{2}+xc+c^{2} is not too large.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Opalg View Post
    For (a), \delta=\varepsilon doesn't work. You need to take \delta=\varepsilon^3. That will ensure that \delta^{1/3}=\varepsilon.

    For (b), the idea is this. You want to show that if |x-c| is small enough then |x^3-c^3| will be small. But |x^{3}-c^{3}|=|x-c||x^{2}+xc+c^{2}|. So you have to concentrate on ensuring that x^{2}+xc+c^{2} is not too large.
    So, I want solve this

    \delta=\frac{\epsilon}{|x^{2}+xc+c^{2}|}

    so that |x-c|<\frac{\epsilon}{|x^{2}+xc+c^{2}|}

    Thus, I want to bound |x^{2}+xc+c^{2}|

    correct?

    If so, I am still getting stuck on bounding that term.

    Can a proof like this be workded in this way:

    c-\delta<c<c+\delta (the delta neighborhood centered at c)
    .
    .
    .
    f(c)-\epsilon<f(c)<f(c)+\epsilon

    ???
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    So, I want solve this

    \delta=\frac{\epsilon}{|x^{2}+xc+c^{2}|}

    so that |x-c|<\frac{\epsilon}{|x^{2}+xc+c^{2}|}

    Thus, I want to bound |x^{2}+xc+c^{2}|

    correct?

    If so, I am still getting stuck on bounding that term.

    Can a proof like this be worked in this way:

    c-\delta<c<c+\delta (the delta neighborhood centered at c)
    .
    .
    .
    f(c)-\epsilon<f(c)<f(c)+\epsilon

    ???
    There's a little trick that helps in calculations like this. Let's agree from the start that we are going to choose \delta < |c| (but we'll keep open the option of making futher restrictions on \delta later on).

    Then if |x-c|<\delta < |c|, it follows that |x| must lie between 0 and 2|c|. In particular, |x|<2|c|. So |x^{2}+xc+c^{2}|\leqslant |x^{2}|+|xc|+|c^{2}|< 4|c|^2 + 2|c|^2 + |c|^2 = 7|c|^2. This tells us that |x^3-c^3| = |x-c||x^{2}+xc+c^{2}| < 7c^2\delta.

    Now we make a further restriction on \delta, namely \delta <\varepsilon/(7|c|^2). That will ensure that |x^3-c^3| < \varepsilon.

    Putting it all together, the way to choose \delta is to require that 0<\delta<\min\{|c|, \varepsilon/(7|c|^2)\}.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Opalg View Post
    There's a little trick that helps in calculations like this. Let's agree from the start that we are going to choose \delta < |c| (but we'll keep open the option of making futher restrictions on \delta later on).

    Then if |x-c|<\delta < |c|, it follows that |x| must lie between 0 and 2|c|. In particular, |x|<2|c|. So |x^{2}+xc+c^{2}|\leqslant |x^{2}|+|xc|+|c^{2}|< 4|c|^2 + 2|c|^2 + |c|^2 = 7|c|^2. This tells us that |x^3-c^3| = |x-c||x^{2}+xc+c^{2}| < 7c^2\delta.

    Now we make a further restriction on \delta, namely \delta <\varepsilon/(7|c|^2). That will ensure that |x^3-c^3| < \varepsilon.

    Putting it all together, the way to choose \delta is to require that 0<\delta<\min\{|c|, \varepsilon/(7|c|^2)\}.
    Sorry, I have one more question about the final proof. So, |g(x)-g(c)|=|x^{1/3}-c^{1/3}|<|x^{3}-c^{3}|=<br />
|x-c||x^{2}+xc+c^{2}|<br />
<\frac{\epsilon}{7|c|^{2}}7|c|^{2}=\epsilon

    How do I relate |g(x)-g(c)|=|x^{1/3}-c^{1/3}| to |x^{3}-c^{3}| in the proof? My teacher gave us the identity a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) and you showed how it is used in the proof, but where do I introduce it?

    Also, when is it best to use the "trick" you showed above?
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  6. #6
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    Quote Originally Posted by Danneedshelp View Post
    Sorry, I have one more question about the final proof. So, |g(x)-g(c)|=|x^{1/3}-c^{1/3}|<|x^{3}-c^{3}|=<br />
|x-c||x^{2}+xc+c^{2}|<br />
<\frac{\epsilon}{7|c|^{2}}7|c|^{2}=\epsilon

    How do I relate |g(x)-g(c)|=|x^{1/3}-c^{1/3}| to |x^{3}-c^{3}| in the proof? My teacher gave us the identity a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}) and you showed how it is used in the proof, but where do I introduce it?
    In fact, my previous comment was completely misdirected. It showed how to prove the continuity of the function x^3, not x^{1/3}. So let's try again.

    You want to show that x^{1/3} is continuous at x=c, where c≠0, and you have the identity a^3-b^3 = (a-b)(a^2+ab+b^2). Apply the identity with a=x^{1/3} and b = c^{1/3}. That gives x-c = (x^{1/3}-c^{1/3})(x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}), from which |x^{1/3}-c^{1/3}| = \frac{|x-c|}{|x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}|}. We can make the numerator of that fraction small, so we must ensure that the denominator does not also get too small. Therefore we want a lower bound on the size of the denominator.

    Going back to the expression a^2+ab+b^2, if a and b have the same sign then all three terms in that expression are positive, and so a^2+ab+b^2 > b^2. Therefore, if we can ensure that x and c have the same sign, it will follow that x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}> c^{2/3}. But we can ensure that x and c have the same sign, by requiring that \delta<|c| (because that tells us that |x-c|<|c|).

    Thus if \delta<|c| we know that |x^{1/3}-c^{1/3}| = \frac{|x-c|}{|x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}|} < \frac{|x-c|}{c^{2/3}}. If we also knew that \delta<c^{2/3}\varepsilon then it would follow that |x^{1/3}-c^{1/3}| < \varepsilon, as required.

    So the way to choose \delta is to require that 0 < \delta < \min\{|c|,c^{2/3}\varepsilon\}.
    Last edited by Opalg; November 9th 2009 at 11:28 AM.
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