Epsilon-Delta proof for continuity 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?
 A: 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.
A: 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 $|f(x) - f(c)| < \varepsilon$.
