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i'm trying to learn the concept of continuity in a more or less formal basis. One definition of continuity (not the most general, but I better learn step by step) is that a function

$$f:X\subseteq\mathbb{R}\to Y\subseteq\mathbb{R}$$

is continuous at $x_0\in X$ if $$\lim_{x\to x_0}f(x)=f(x_0)$$

This is, $\forall \epsilon>0, \exists \delta>0$ such that $|f(x)-f(x_0)|<\epsilon$ if $x\in X$, $0<|x-x_0|<\delta$

So in order to test if I do understand this, decided to try a particular example, say $f(x)=x^2$, and $f(x_0)=x_0^2$. My first question is:

1) Is an assumption that $f(x_0)=x_0^2$ or do I need to prove it? I think I don't have to prove it because that equality follows from the definition of $f(x)$ but there may be any kind of subtlety that I don't see.

Assuming I can say $f(x_0)=x_0^2$ then the task is to show that there exists such a number $\delta$:

I'm given $\epsilon>0$ such that $|f(x)-f(x_0)|<\epsilon$ this is $|x^2-x_0^2|<\epsilon$, Rewriting $|x^2-x_0^2|=|x-x_0||x+x_0|<\epsilon$

2) Can I set


what if $x=-x_0$?

Thanks for your time.

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Nitpick: in order for your definition to make sense, $x_0$ needs to be a limit point of $X$. – wj32 Feb 22 '13 at 12:22
@wj32 Does that $x_0$ is a limit point of $X$ means that every point of $X$ is not $x_0$? But does this not prevent it to be different from $-x_0$ (if what I say its right) – Jorge Feb 22 '13 at 12:31
@wj32: no, you don't really need to... this would complicate the definition of continuity (because you should assert that $f$ is continuous on insulated points) – Emanuele Paolini Feb 22 '13 at 12:34
@Nivalth: Consider any function $f:\mathbb{Z}\to\mathbb{R}$. Such a function is always continuous, but it does not make sense to talk about the limit of $f(x)$ as $x \to n$. My point is that the definition of continuity that you have provided is incomplete. Of course, this doesn't affect the rest of your question, so don't worry about it too much. See – wj32 Feb 22 '13 at 12:35
up vote 1 down vote accepted

1) $f(x_0)$ is given, in this case $f(x_0)=x_0^2$. To show continuity of $f$, we have to show $$ \lim_{x\to x_0}f(x)=f(x_0) $$ 2) $\delta$ determines which $x$ are in $|x-x_0|\le\delta$. It is not proper to have $\delta$ depend on $x$; it should only depend on $x_0$ and $\epsilon$.

For example, if $\delta<\min\left(\dfrac{\epsilon}{3x_0}\,,\,\dfrac{x_0}{2}\right)$, then since $x+x_0=2x_0+(x-x_0)$, we have $$ \begin{align} \left|\,x^2-x_0^2\,\right| &=\left|\,x-x_0\,\right|\left|\,x+x_0\,\right|\\ &\le\dfrac{\epsilon}{3x_0}\frac{5x_0}{2}\\ &=\frac{5\epsilon}{6}\\ &\le\epsilon \end{align} $$

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Thanks for the edit, and the answer. So the point is to find a number $\delta=\delta(\epsilon,x_0)$ such that $0<|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\epsilon$ ? Or I'm missing it one more time – Jorge Feb 22 '13 at 12:56
@Nivalth: that is precisely it :-) – robjohn Feb 22 '13 at 13:16

There is a fundamental misunderstanding here what the hypothesis is and what you have to prove. The square function is continuous in $x_0$ means (in words): for every positive $\varepsilon$ there exists a positive number $\delta$ such that for every number $x$ with the property $|x-x_0|<\delta$, the inequality $|x^2-x_0^2|<\varepsilon$ holds.

So you are not given an $\varepsilon$ with $|x^2-x_0^2|<\varepsilon$, rather, that is what you ultimately have to prove.

Maybe the square function is not the easiest example to start with; maybe it helps to try a linear function first ($f(x)=ax+b$).

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Hi thanks, but I don't get it yet. So what I'm give is the $\delta$ part instead? So, for the linear function example I pick up any number $x$ such that $|x-x_0|< \delta$ for any $\delta >0$ and now I compute $|f(x)-f(x_0)|=|ax+b-(ax_0+b)|=|a(x-x_0)|=|a||x-x_0| $,how does the $\epsilon$ appears now? – Jorge Feb 22 '13 at 12:41
Well I have $|f(x)-f(x_0)|=|a||x-x_0|<|a|\delta$ so $\epsilon < |a|\delta$ ? – Jorge Feb 22 '13 at 12:43
If you have to prove that for all $\varepsilon$, something holds, then you are given $\varepsilon$ and prove that something. If you have to prove that there exists $\delta$ such that something holds, then you can choose one $\delta$ you like, and prove it only with this one. – Florian Feb 22 '13 at 13:18
In this particular case (linear function) it may be $\delta=\frac{\epsilon}{|a|}$ – Jorge Feb 22 '13 at 13:22
  1. You have defined $f(x)$ so that $f(x_0)=x_0^2$, so this is granted. But you could have defined $f$ differently...
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