The proper limit at the non-proper point

The limit of a function is one of the most important concepts in mathematical analysis. It describes the behaviour of a function around a certain point, allowing us to define, for example, the continuity of a function. The limit of a function helps us understand the behaviour of a function even at points where it is not defined at all.

The proper limit at a non-proper point

This situation is similar to the limits of a sequence. We are looking for what the limit of a function is when we approach infinity. The limit can be either proper or non-proper. We start with the proper limit.

Let ∞ be the bulk point of the function f. Then L ∈ ℝ is the limit of the function at the point ∞, if

$$(\forall\epsilon>0),(\exists A\in \mathbb{R}),(\forall x\in D(f)),(x > A \Rightarrow |f(x)-L|<\epsilon)$$

and at the point −∞ if

$$(\forall\epsilon>0),(\exists A\in \mathbb{R}),(\forall x\in D(f)),(x < A \Rightarrow |f(x)-L|<\epsilon)$$

What does the definition tell us? That if we choose some ε-environment around our intended limit L (i.e., on the y axis), then we are always able to find a point A on the x axis such that if we take any point x to the right of A, i.e., x>A, then |f(x)−L| < ε will hold, i.e., all function values will be less than ε away from the limit L.

Consider the function f(x) = (2/x)+1. Graph:

It can be seen that as the value of x approaches infinity, the functional value of f(x) approaches one. So we choose L = 1. Now we prove that L = 1 is indeed the limit of this function at plus infinity. First, let's try it out. We will choose some epsilon, for example $\epsilon=\frac12$. Now we'll try to find a A ∈ ℝ such that it holds that for all x that are greater than A, the function value is less than ε away from L. Let's sketch this in the figure:

That the functional value f(x) is less away from L than ε means that the curve describing the graph of the function is between those red lines. We must now find the boundary A on the axis x, from which this condition is satisfied. This need not necessarily be the smallest possible boundary, so we can choose, for example, A = 6.

We see that the curve describing the graph beyond the A limit is entirely between the red lines that mark the ε-distance from L. Now we have proved this for one particular ε, to prove that L = 1 is indeed the limit, we need to be able to prove it for all ε > 0.

Now we need to see for which A this relation holds:

$$x > A \Rightarrow |f(x)-L| < \epsilon$$

After reaching:

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

We can subtract ones:

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

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

$$x > A \Rightarrow \frac{2}{x} < \epsilon$$

We multiply x:

$$x > A \Rightarrow 2 < x\cdot\epsilon$$

Divide by ε:

$$x > A \Rightarrow \frac{2}{\epsilon} < x$$

By adjusting, we have found that x must be greater than 2/ε. So if we choose some number after A that is greater than 2/ε, then we find the limit we were looking for.

In the previous experiment we chose $\epsilon=\frac12$, so we should choose a number after A that is greater than $2/\frac12=4$. We chose A = 6, which is fine.