Dtjytyk

I'm studying for an analysis test.

The definition that I have been given from class seems pretty standard:

$$left | x-c right | leq delta Rightarrow left | f(x)-f(c) right |leq varepsilon$$

But it has the added stipulation: for all $x$ in the domain of $f$. I went back to the slides we had in class and checked that I didn't copy this down wrong.

This part is confusing me. If I were proving that $y = x$ is continuous at $c = 0$ then for $ left | x-c right | $ I can always find an $x$ in the domain of $f$ that is larger than the chosen delta. Our domain is infinite so then our delta would be infinite. Is the problem that I am thinking of delta as a specific value when it doesn't need to be a named value?

Sorry if my question is not concise. Any edits are appreciated.

I'm not which part of the definition specifically confuses you, so I included a long answer that hopefully clears any confusions you had.

The reason we add that $x$ needs to be in the domain of $f$ is because we want to make sure that the expression $f(x)$ makes sense. The domain of $f$ is, essentially, all the points $x$ such that $f(x)$ is defined (i.e. makes sense).

Consider for instance the function $f(x) = 1/x$. This function is not defined at $0$ and it therefore does not make sense to write $f(0)$.

As for the $epsilon-delta$ definition, let's begin with trying to understand it conceptually. For a function to be continuous at a point $c$, what do we want? We want to make sure that we can always get "infinitely close" to the value $f(c)$ when $x$ is "close enough to $c$. Mathematically, we say that for any $epsilon >0$, it is possible to find a number $delta>0$ such that whenever $x$ is sufficiently close to $c$ (i.e. $lvert x-c rvert < delta$), our function $f$ is close to the value $f(c)$ (i.e. $lvert f(x) - f(c) rvert <epsilon$. Of course, we assume that $x$ is the the domain of $f$ since the expression $f(x)$ does not make sense otherwise.

Let's now go through an example of how to apply this mathematical definition. Suppose you are given a function $f$ which is defined only at the point $0$, say $f(0) = 1$. Then the domain of $f$ is the point $0$ which means that it only makes sense to write $f(x)$ if $x=0$. Now, using our $epsilon-delta$ definition to show that $f$ is continuous at $c=0$ is not too hard;

Let $epsilon>0$ be an arbitrary positive number (recall that our definition required our inequalities to hold for any $epsilon >0$. We now pick $delta = 1$ (we need to make sure $delta > 0$ exists, it can be any number $>0$ - you just pick $delta>0$ in a way that ensures $lvert x-crvert <delta implies lvert f(x)-f(c)rvert < epsilon$). Then for any $x$ (i.e. $x=0$) in the domain of $f$ with
$$
lvert x-c rvert < delta = 1,
$$
we have
$$
lvert f(x) - f(c) rvert = lvert f(0) - f(0) rvert = 0 < epsilon
$$
as desired. Note that the first equality above holds since $x=0$ and $c=0$.

The above proof is relatively easy as long as you can wrap your mind around the definition. I'm really not doing much in the proof. It's obvious that if $f$ is only defined at one point than it is continuous - I'm just directly using the definition to prove it.

For extra practice, let's now prove that the function $f(x) = x$, which is defined on all real numbers, is continuous at any point $c$.

For any $epsilon >0$, we pick $delta = epsilon> 0$. Then for any $x$ such that $lvert x-crvert < delta$, we see that
$$
lvert f(x) - f(c) rvert = lvert x-crvert < delta = epsilon.
$$
So for any number $epsilon >0$, we have found $delta>0$ such that
$$
lvert x-crvert < delta implies lvert f(x) - f(c) rvert < epsilon
$$
which is what had to be shown.