I have a lot of trouble figuring out how to work with this proof technique for continuity. The definition says:

$$ \forall \varepsilon \space \exists \delta \quad |x-a|\lt \delta \quad \Rightarrow \quad |f(x)-f(a)| \lt \varepsilon$$

That means that, in order to have a continuous function, there should exist a $\delta$, for every $\varepsilon$ we are given, that satisfies the implication above, right?

So which would be the first steps? I mean, sometimes I see proofs that starts immediately by choosing the possible $\delta$. But I think they first calculated to find the values, and then started the proof in that way right (proof structure I guess)?

To make an example, how would you show that $f:[0,\Re] \rightarrow[0,\Re], \space f(x)= \sqrt(x)$ is continuous?

$$|\sqrt{x}- \sqrt{a}|= \lvert\frac{x-a}{\sqrt{x}+\sqrt{a}}\rvert \lt |\frac{\delta}{\sqrt{x}+\sqrt{a}}|\lt |\frac{\delta}{\sqrt{a}}| = \varepsilon$$ for $\delta = \varepsilon \sqrt{a}$.

Are the steps right? For example the fact that I can simplify by deleting $\sqrt{x}$? Which would be the conclusion in order to have a formal proof?

In this example we manage to get the $\delta$ easily, but how would you proceed if there is no way to get there?

  • $\begingroup$ A common technique is to leave a gap near the beginning of the proof where you have to choose a $\delta$, then work your way through the proof and then pick $\delta$ when it becomes obvious what it should be. Can you give an example of where $\delta$ may be "hard" to choose? $\endgroup$ – Irregular User Nov 24 '15 at 18:29
  • $\begingroup$ There is something else to keep in mind when first learning this material: this is a wonderful formal definition of a concept (thankyou Cauchy) that can be used to develop a deep theory. You, however, as a lowly student are expected to drill on how to use it to prove continuity in specific cases. This is just the drudge work of trying to get familiar with the concept. The end result of this, far down the road that you cannot yet see, is that you will have sufficient skills to handle the theory. Then the techniques mentioned here will seem kind of quaint. $\endgroup$ – B. S. Thomson Nov 24 '15 at 18:56
  • $\begingroup$ @IrregularUser For example to prove the continuity $\endgroup$ – Ergo Nov 25 '15 at 14:50
  • $\begingroup$ @Ergo I mean, if you have a specific example of a continuity proof where it is difficult to choose \delta, then we could walk you through how to prove continuity and pick \delta. $\endgroup$ – Irregular User Nov 25 '15 at 16:32
  • $\begingroup$ @IrregularUser Well for example a trigonometric function, let's say $f(x)=sin(x)$ $\endgroup$ – Ergo Nov 25 '15 at 21:57

There is no "sure fire" way of proving continuity of a function. However, the steps are usually a bit backward to what the actual definition is. That is, the definition says that $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta >0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

We start the proof by taking an arbitrary $\epsilon > 0$. However, we then usually do not magically think of a $\delta$ that would fit. What we typically try to do is simplify the expression $|f(x)-f(a)|$ and prove that it is "small", keeping in the back of our mind that we can always make $|x-a|$ "small".

In your case, for example, we can prove first that $|f(x)-f(a)|=|\frac{x-a}{\sqrt x + \sqrt a}|$. Now, we want to ask ourselves: If $|x-a|$ is small, is this expression also small?

This is done by trying to find a small upper bound for the expression that will hold whenever $|x-a|<\delta$. In your particular case, this is fairly simple, since we know that the expression is smaller than $\frac{|x-a|}{\sqrt a}$, and this is smaller than $\epsilon$ if $\delta$ is set small enough.

Now, once we did these steps, we take a step back, and think about what we just did.

Looks like we found our $\delta$, and we have to set our $\delta=\epsilon\sqrt a$.

Hmm, is this OK? Well, as long as $a>0$, we know that $\delta>0$, and we have proven that if $|x-a|<\delta$, then $|f(x)-f(a)|<\frac{\delta}{\sqrt a}$, ,which is equal to $\epsilon$, so it looks like we are almost done.

Almost, because we can see that if $\sqrt a=0$, then our proof does not work! We need to do that part separately. If $a=0$, then $|f(x)-f(a)|=|\sqrt x|$. We still need to prove that if $|x-0|$ is small, then $|\sqrt x|$ is also small.

  • $\begingroup$ Thanks for the advices, in the last case $\delta$ would be $\varepsilon^2$ right? $\endgroup$ – Ergo Nov 24 '15 at 18:54
  • $\begingroup$ @Ergo Yup. And 2 lines of a proof. $\endgroup$ – 5xum Nov 24 '15 at 18:59

Let me show you a succinct argument, which illustrates:

Let $c \geq 0$. If $x \geq 0$, then $$ |\sqrt{x} - \sqrt{c}| = \frac{|x-c|}{\sqrt{x} + \sqrt{c}} \leq \frac{|x-c|}{\sqrt{c}}; $$ given any $\varepsilon > 0$, we have $|x-c|/\sqrt{c} < \varepsilon$ if in addition $|x-c| < \varepsilon \sqrt{c}$. Hence, for every $c \geq 0$ and every $\varepsilon > 0$ it holds that $x \geq 0$ and $|x-c| < \varepsilon \sqrt{c}$ imply $|f(x) - f(c)| < \varepsilon$.


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