Here we will show how one goes about proving the following statement.

Let f, g be functions defined on a neighborhood of some point a. If f and g are continuous at a, then the function f + g is also continuous at a.

What kind of proof will we use? At the first look there is no reason to fear a direct proof. Being continuous is a definite information, so this may be a good beginning. On the other hand, not being continuous is a complicated information (because there are many ways in which continuity can go wrong), so passing to negation does not seem very promising. This suggests that an indirect proof or a proof by contradiction are not a good idea. The conclusion is that we try a direct proof.

How does one go about proving that f + g is continuous at a? The first step is usually to look at definition. Here there are two possibilities. Some author first develop the notion of limit and define continuity as follows:

Definition.
A function h is continuous at a if and only if the limit of h at a is equal to h (a).

We are ready to start the proof. Let us take some point a and two functions f, g defined on some neighborhood of a. We will also assume that they are continuous at a, which means that the above condition also applies to f and g. Now we need to show that the sum f + g is a continuous function at a.

What do we know about the limit of f + g at a? When investigating limits there was surely a theorem that the limit of a sum f + g at some point is equal to the limit of f plus the limit of g there.

lim_x→a( f + g) = lim_x→a( f ) + lim_x→a(g).

Now do we know anything about those two limits on the right? Actually, yes, we assumed that f and g are continuous at a, so we have

lim_x→a( f ) + lim_x→a(g) = f (a) + g(a).

When we put it together, we get

lim_x→a( f + g) =  lim_x→a( f ) + lim_x→a(g)       [by the teorem on linearity of a limit]
      = f (a) + g(a)       [by continuity of f and g at a]
      = ( f + g)(a).

By definition, this proves that the function f + g is indeed continuous at a.

Note how every step in the proof was supported by some argument. Here the main step was supported by a theorem that was proved about limits. This is typical, in mathematics we always try to use what we already did before in order to save work.

However, other authors do continuity without using limit. Their definition then goes like this:

Definition.
A function h is continuous at a when the following is true: For every e > 0 there exists d > 0 such that for all real numbers x: if |x − a| < d, then |h(x) − h(a)| < e.

Note that we did not use the traditional epsilon and delta there. We had a good reason, it gives that definition a certain flexibility that we will need later. It also shows that particular choice of letters is not important, the crucial thing is the meaning. What does the definition say?

We will try to translate symbols into meanings. Given a certain quantity, the tolerance, we need to find how far we can go away from a, which is specified by another quantity, without values of h getting too far (as measured by tolerance) from h(a). In other words, somebody specifies a tolerance and we have to force values of h to stay close (as specified by this tolerance) to h(a), we do this forcing by restricting the movement of x away from a. Obviously there si a dependence, when somebody gives us a really small tolerance for values of h, then we can expect that also the freedom of movement of x will get restricted in order to keep values of h where they should be.

Now we apply this definition to our situation. We want to do the direct proof, which means that we want to show that the sum f + g is continuous at a, that is, we want to prove that the following is true:

For every ε > 0 there exists δ > 0 such that for all real numbers x: if |x − a| < δ, then |( f + g)(x) − ( f + g)(a)| < ε.

This starts with a general quantifier, so in order to show that this whole statement is true, we have to start like this: Let ε be an arbitrary positive number. We need to show that the rest of the statement is true, that is, we have to find some delta such that the rest is true. In order to find this delta we first need to understand the condition. What happened? Somebody gave us the tolerance ε, we do not know what it is (a general quantifier) but we do have it now. We need to force the values of f + g to stay near the number ( f + g)(a) as long as we do not stray far from a, we need to find how big is that "far". What tools do we have for it? Well, we know that both f and g are continuous at a.

This is the crucial moment, we need to find some bridge between what we have and what we need to get, first on a conceptual level. Here it seems quite obvious. The fact that f and g are continuous at a means that values of f stay near f (a) and values of g stay near g(a) as long as we do not get far from a with x. Intuitively it would seem that then also the sum of values should stay close to the sum of the target values. It looks like we did find the right bridge, but now there is the problem of doing it formally, that is, writing the proper mathematics about it. We start by formalizing (and checking) our guess that if f and g are close etc, then the same works for the sum. We need to look at the difference in absolute value and hypnotize it for a while, thinking that we somehow need this to reflect individual information about f and g, and sooner or later the following should appear in our mind.

|( f + g)(x) − ( f + g)(a)| = | f (x) + g(x) − f (a) − g(a)|
= |[ f (x) − f (a)] + [g(x) − g(a)]|.

This looks promising. Is there a way to actually break the part with f away from the part with g? Yes, the so-called triangle inequality is exactly what we need here.

|( f + g)(x) − ( f + g)(a)| ≤ | f (x) − f (a)| + |g(x) − g(a)|.     (*)

This is a tricky moment, since inequality always works in just one direction. Are we lucky, is it the direction we want? By our assumption, we can make values of f close to f (a) and similarly for g, therefore we can make the two expressions on the right small, the inequality then forces also the expression on the left to be small. Yes, on this level things seem to work, another indication that we are on the right track. Now it is time to get to the real thing. We need to make the expression on the left smaller than ε, which is some number that we do not know, but it is a concrete number that somebody gave us. We know that we have power over the expressions on the right. Ater some thinking you hopefully get this wonderful idea: If we make each expression on the right less than ε/2, then we are essentially done. We do the forcing on the right by restricting movement of x close to a, which is something we are expected to do anyway, so no problem here. We might be ready to write the formal proof.

Proof: Let ε be an arbitrary positive number.

1a. Since f is continuous at a (one of our basic assumptions), the condition in the definition of continuity is true for this f and a. Therefore we can apply it with tolerance e = ε/2 to obtain a certain d_f > 0 such that all x satisfying |x − a| < d_f also satisfy | f (x) − f (a)| < ε/2.

1b. Since g is continuous at a (one of our basic assumptions), the condition in the definition of continuity is true for this g and a. Therefore we can apply it with tolerance e = ε/2 to obtain a certain d_g > 0 such that all x satisfying |x − a| < d_g also satisfy |g(x) − g(a)| < ε/2.

Intermezzo: In order to make theleft hand-side in (*) small, we need to keep both parts on the right small. This means that both restrictions for x derived in steps 1a. and 1b. must apply simultaneously. Thus the distance that x is allowed to move must be chosen in such a way that x does not move further than given by the two d's above, that is, we need to take the smaller restriction of the two.

2. Define δ = min(d_f,d_g). Then δ > 0 and we claim that this delta satisfies the condition from definition of continuity.

Indeed, take any x such that |x − a| < δ. By the definition of delta this means that also |x − a| < δ ≤ d_f, hence by 1a, | f (x) − f (a)| < ε/2. Similarly, |x − a| < δ ≤ d_g, hence by 1b, | g(x) − g(a)| < ε/2. Therefore

|( f + g)(x) − ( f + g)(a)| ≤ | f (x) − f (a)| + |g(x) − g(a)| < ε/2 + ε/2 = ε,

exactly as needed. The proof is complete.

This second proof was much harder than the first. However, this does not mean that people who follow the first way have it easier. In fact, they also have to do this messy work with inequalities and epsilons, but they have to do it when proving the theorem on linearity of a limit. So one cannot really avoid it, but it makes sense to do it for limits first, since one has to do it eventually anyway and if we do it first, then we can use this result to get a nice proof for continuity as above.