Finite limit at infinity

The limit of a function is one of the most important ideas in mathematical analysis. It describes how a function behaves near a particular point, which lets us define, for example, the continuity of a function. Limits help us understand how a function behaves even at places where it is not defined at all.

Finite limit at infinity

This situation is similar to limits of sequences. We look for the limit of a function as we approach infinity. The limit can be either finite or infinite. We start with a finite limit.

Let ∞ be an accumulation point of dom(f), the domain of the function. Then L ∈ ℝ is the limit of the function at the point ∞ if

$$(\mathrm{\forall }ϵ>0)(\mathrm{\exists }A\in \mathbb{R})(\mathrm{\forall }x\in \mathrm{dom}(f))(x>A\Rightarrow |f(x)-L|<ϵ)$$

and at the point −∞ if

$$(\mathrm{\forall }ϵ>0)(\mathrm{\exists }A\in \mathbb{R})(\mathrm{\forall }x\in \mathrm{dom}(f))(x<A\Rightarrow |f(x)-L|<ϵ)$$

What does the definition tell us? That if we choose some ϵ-neighbourhood around the limit L we have in mind (that is, on the y-axis), then we can always find a point A on the x-axis such that for any point x to the right of A, that is x>A, we have |f(x) − L| < ϵ, so all the function values are less than ϵ away from the limit L.

Let’s take the function f(x) = (2/x) + 1. The graph:

You can see that as the value of x approaches infinity, the function value f(x) approaches one. So we choose L = 1. Now we will prove that L = 1 really is the limit of this function at plus infinity. First we try it out. We choose some epsilon, for example $ϵ=\frac{1}{2}$. Now we try to find an A ∈ ℝ such that for all x greater than A, the function value is less than ϵ away from L. Let’s sketch it in a picture:

The function value f(x) being less than ϵ away from the point L means that the curve describing the graph of the function lies between those red lines. We now have to find a bound A on the x-axis from which this condition is satisfied. It doesn’t have to be the smallest possible bound, so we can choose, for example, A = 6.

We can see that beyond the bound A the curve describing the graph lies entirely between the red lines, which mark the ϵ-distance from L. So far we have shown this for one particular ϵ. To prove that L = 1 really is the limit, we must be able to prove it for all ϵ > 0.

Now we have to find for which A this relationship holds:

$$x>A\Rightarrow |f(x)-L|<ϵ$$

After substitution:

$$x>A\Rightarrow |(\frac{2}{x}+1)-1|<ϵ.$$

We can subtract the ones:

$$x>A\Rightarrow \left|\frac{2}{x}\right|<ϵ$$

Next we can assume that we only take positive x (we are interested in values of x approaching plus infinity), which lets us remove the absolute value:

$$x>A\Rightarrow \frac{2}{x}<ϵ$$

We multiply by x:

$$x>A\Rightarrow 2<x\cdot ϵ$$

We divide by ϵ:

$$x>A\Rightarrow \frac{2}{ϵ}<x$$

Rearranging, we have found that x must be greater than 2/ϵ. So if we choose for A any number greater than 2/ϵ, we find the bound we were looking for.

In the previous attempt we chose $ϵ=\frac{1}{2}$, so we should choose for A a number greater than $2/\frac{1}{2}=4$. We chose A = 6, which is fine.