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Thread: Continuity proof

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

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

    Q:

    a) Prove that g is continuous at c=0

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

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

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

    So, if $\displaystyle \delta=\epsilon$, then $\displaystyle |x|<\delta$ implies $\displaystyle |x^{1/3}|<\epsilon$. Therefore, $\displaystyle g(x)$ 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 $\displaystyle g(x)=x^{1/3}$

    Q:

    a) Prove that g is continuous at c=0

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

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

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

    So, if $\displaystyle \delta=\epsilon$, then $\displaystyle |x|<\delta$ implies $\displaystyle |x^{1/3}|<\epsilon$. Therefore, $\displaystyle g(x)$ 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), $\displaystyle \delta=\varepsilon$ doesn't work. You need to take $\displaystyle \delta=\varepsilon^3$. That will ensure that $\displaystyle \delta^{1/3}=\varepsilon$.

    For (b), the idea is this. You want to show that if $\displaystyle |x-c|$ is small enough then $\displaystyle |x^3-c^3|$ will be small. But $\displaystyle |x^{3}-c^{3}|=|x-c||x^{2}+xc+c^{2}|$. So you have to concentrate on ensuring that $\displaystyle x^{2}+xc+c^{2}$ 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), $\displaystyle \delta=\varepsilon$ doesn't work. You need to take $\displaystyle \delta=\varepsilon^3$. That will ensure that $\displaystyle \delta^{1/3}=\varepsilon$.

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

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

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

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

    correct?

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

    Can a proof like this be workded in this way:

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

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

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

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

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

    correct?

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

    Can a proof like this be worked in this way:

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

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

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

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

    Putting it all together, the way to choose $\displaystyle \delta$ is to require that $\displaystyle 0<\delta<\min\{|c|, \varepsilon/(7|c|^2)\}$.
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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 $\displaystyle \delta < |c|$ (but we'll keep open the option of making futher restrictions on $\displaystyle \delta$ later on).

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

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

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

    How do I relate $\displaystyle |g(x)-g(c)|=|x^{1/3}-c^{1/3}|$ to $\displaystyle |x^{3}-c^{3}|$ in the proof? My teacher gave us the identity $\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ 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, $\displaystyle |g(x)-g(c)|=|x^{1/3}-c^{1/3}|<|x^{3}-c^{3}|=
    |x-c||x^{2}+xc+c^{2}|
    <\frac{\epsilon}{7|c|^{2}}7|c|^{2}=\epsilon$

    How do I relate $\displaystyle |g(x)-g(c)|=|x^{1/3}-c^{1/3}|$ to $\displaystyle |x^{3}-c^{3}|$ in the proof? My teacher gave us the identity $\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ 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 $\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)$. Apply the identity with $\displaystyle a=x^{1/3}$ and $\displaystyle b = c^{1/3}$. That gives $\displaystyle x-c = (x^{1/3}-c^{1/3})(x^{2/3}+x^{1/3}c^{1/3}+c^{2/3})$, from which $\displaystyle |x^{1/3}-c^{1/3}| = \frac{|x-c|}{|x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}|}$. 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 $\displaystyle a^2+ab+b^2$, if a and b have the same sign then all three terms in that expression are positive, and so $\displaystyle a^2+ab+b^2 > b^2$. Therefore, if we can ensure that x and c have the same sign, it will follow that $\displaystyle x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}> c^{2/3}$. But we can ensure that x and c have the same sign, by requiring that $\displaystyle \delta<|c|$ (because that tells us that |x-c|<|c|).

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

    So the way to choose $\displaystyle \delta$ is to require that $\displaystyle 0 < \delta < \min\{|c|,c^{2/3}\varepsilon\}$.
    Last edited by Opalg; Nov 9th 2009 at 10:28 AM.
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