We want to prove that f (x) = 2x − 1 converges to 5 as x goes to 3. To this end we have to show that we can win the game from the definition.

Assume we were given a positive ε. We need to find a δ > 0 so that if 0 < |x − 3| < δ, then | f (x) − 5| < ε.

Typical approach is to look closer at the inequality we are trying to prove and somehow work the expression |x − 3| into it. Here it is rather easy.

| f (x) − 5| < ε
|(2x − 1) − 5| < ε
|2x − 6| < ε
|2(x − 3)| < ε
2|x − 3| < ε
|x − 3| < ε/2.

Note that the operations were all equivalent, so the first and last lines are equivalent. Now let's review it. We want to make the first line true, and what we can do is to make |x − 3| as small as we wish by choosing the right δ. Now it seems clear what we should do: We choose δ = ε/2. We can do it, since ε was already given at the beginning, we do not know its value but it is given and fixed and so we can use it.

Now we check that what we did fulfills the requirement from the definition. Somebody gave us an arbitrary ε > 0 (that "arbitrary" is the key here, we did it for all epsilons, not just a certain nice one). We then decided to choose δ = ε/2. Is this the right delta? Let's check. Any x that satisfies 0 < |x − 3| < δ then also satisfies |x − 3| < ε/2 and therefore (we reverse the above steps) | f (x) − 5| < ε, exactly as needed, we won the game.

The proof is complete.