We want to prove that f (x) = 2x − 1 goes to infinity as x goes to infinity. To this end we have to show that we can win the game from the definition. We will use the general one with neighborhoods.

Assume we were given some ε > 0. We need to find a δ > 0 so that if x∈U_δ(∞), then also f (x)∈U_ε(∞).

We start by translating this into inequalities. We need to make sure that f (x) > 1/ε and we do it by restricting x to the neighborhood of infinity given by the inequality x > 1/δ. We start by exploring the desired inequality.

f (x) > 1/ε
2x − 1 > 1/ε
2x > 1/ε + 1
x > (1/ε + 1)/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 as large as we wish by choosing the right δ. Now it seems clear what we should do: We choose δ = 2/(1/ε + 1).

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 = 2/(1/ε + 1). Is this the right delta? Let's check. Any x that satisfies x∈U_δ(∞) then by definition of this neighborhood satisfies x > 1/δ = (1/ε + 1)/2. Therefore x also satisfies f (x) = 2x − 1 > 1/ε, which means that f (x)∈U_ε(∞) exactly as needed, we won the game.

The proof is complete.