Infinite limit at a finite point

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.

Infinite limit at a finite point

In the previous part we talked about a finite limit at a finite point. A point x₀ is finite if x₀ ∈ ℝ, that is, if it is a real number. But by a point we can also mean infinity – such a point is then called a point at infinity. We distinguish plus and minus infinity: +∞, −∞. An infinite limit at a finite point therefore means that we have some function f and a finite point x₀, and the limit at this point turns out to be infinity. An example of such a function is $f(x)=|\frac{1}{x}|$. The graph:

At the finite point x₀ = 0 the function has the limit plus infinity – whichever side we approach x₀ from, the value always grows towards infinity. Let x₀ be an accumulation point of dom(f), the domain of the function. The function f has the limit +∞ at the point x₀ if

$$(\mathrm{\forall }K\in \mathbb{R})(\mathrm{\exists }\delta >0)(\mathrm{\forall }x\in \mathrm{dom}(f))(0<|x-x_0|<\delta \Rightarrow f(x)>K)$$

and −∞ if

$$(\mathrm{\forall }K\in \mathbb{R})(\mathrm{\exists }\delta >0)(\mathrm{\forall }x\in \mathrm{dom}(f))(0<|x-x_0|<\delta \Rightarrow f(x)<K).$$

If the definition is satisfied, we can write

$$\underset{x\to x_0}{lim}f(x)=\pm \mathrm{\infty }.$$

What does the definition tell us? In the first definition the variable K represents a kind of bound on the y-axis. The definition then says that however large (high) a bound K we choose, we can always find some punctured δ-neighbourhood of the point x₀ such that every element of this neighbourhood has a function value greater than K.

As an example, let’s try the definition on the previous function – we will look for the limit at the point x₀ = 0. We choose K = 4:

Now we have to find a value of δ such that all the points in the interval (−δ, 0) ∪ (0, δ) have a function value greater than K. We can choose, for example, δ = 0.2. If we draw this in the picture, we get:

We can see that the function values in the interval (−δ, 0) ∪ (0, δ) lie above the value K. We can also work it out. For x = −0.2 we get f(−0.2) = |−1/0.2| = 5. The function is increasing on the negative numbers, so for all x in the interval (−0.2, 0) the function value will certainly be greater than five, and so also greater than K = 4. Similarly for the second interval.

So far we have only shown this for one particular K. If the function is to have the limit plus infinity, it must hold for any K. Let’s try to prove it. Since f(x) = f(−x) (the function is even), it is enough to work with the positive case only, that is, we restrict ourselves to the interval (0, ∞).

We are given some K and we must find a δ such that f(x) > K for all x ∈ (0, δ). Since the function is decreasing on the positive numbers, it is enough to find a δ for which f(δ) > K. Then it will certainly hold for all x ∈ (0, δ) as well. We now solve the inequality f(δ) > K, which after substitution is:

$$\frac{1}{\delta }>K.$$

Now we multiply the whole inequality by δ. The value δ is always positive by definition, so we are allowed to do this. We get:

$$1>\delta \cdot K$$

Now suppose that K > 0. For K ≤ 0 the proof is trivial, because f(x) > 0 for all x. Now we can divide the whole inequality by K:

$$\frac{1}{K}>\delta .$$

That is the result. If we are given K, we find the value of δ by taking any number that is less than 1/K (and greater than zero, by the definition), and we get our interval (0, δ).

In the previous case we had K = 4. Substituting, we find that 1/4>δ. Delta must be less than one quarter. We chose 0.2, one fifth, and it worked.