# Continuity proof

• Nov 5th 2009, 09:36 PM
Danneedshelp
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
• Nov 6th 2009, 06:09 AM
Opalg
Quote:

Originally Posted by Danneedshelp
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.
• Nov 8th 2009, 11:21 AM
Danneedshelp
Quote:

Originally Posted by Opalg
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)
.
.
.
$\displaystyle f(c)-\epsilon<f(c)<f(c)+\epsilon$

???
• Nov 8th 2009, 12:12 PM
Opalg
Quote:

Originally Posted by Danneedshelp
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 $\displaystyle 0<\delta<\min\{|c|, \varepsilon/(7|c|^2)\}$.
• Nov 8th 2009, 12:49 PM
Danneedshelp
Quote:

Originally Posted by Opalg
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?
• Nov 9th 2009, 08:34 AM
Opalg
Quote:

Originally Posted by Danneedshelp
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 $\displaystyle 0 < \delta < \min\{|c|,c^{2/3}\varepsilon\}$.