# Thread: Continuity proof

1. ## 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

2. Originally Posted by Danneedshelp
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.

3. Originally Posted by Opalg
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 (the delta neighborhood centered at c)
.
.
.
$f(c)-\epsilon

???

4. Originally Posted by Danneedshelp
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 (the delta neighborhood centered at 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 $0<\delta<\min\{|c|, \varepsilon/(7|c|^2)\}$.

5. 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 $\delta < |c|$ (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, $|g(x)-g(c)|=|x^{1/3}-c^{1/3}|<|x^{3}-c^{3}|=
|x-c||x^{2}+xc+c^{2}|
<\frac{\epsilon}{7|c|^{2}}7|c|^{2}=\epsilon$

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?

6. Originally Posted by Danneedshelp
Sorry, I have one more question about the final proof. So, $|g(x)-g(c)|=|x^{1/3}-c^{1/3}|<|x^{3}-c^{3}|=
|x-c||x^{2}+xc+c^{2}|
<\frac{\epsilon}{7|c|^{2}}7|c|^{2}=\epsilon$

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 then it would follow that $|x^{1/3}-c^{1/3}| < \varepsilon$, as required.

So the way to choose $\delta$ is to require that $0 < \delta < \min\{|c|,c^{2/3}\varepsilon\}$.