Limit of a function

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.

Neighbourhood of a point

We begin with the definition of an epsilon neighbourhood of a point in the set of real numbers. Let r ∈ ℝ. Then for ϵ > 0 the epsilon neighbourhood of this point is the open interval (r − ϵ, r + ϵ). The punctured epsilon neighbourhood is the same, except that it does not contain the point r. The punctured neighbourhood is (r − ϵ, r) ∪ (r, r + ϵ). We write the epsilon neighbourhood of a point a as U(a, ϵ) and the punctured neighbourhood as R(a, ϵ).

As an example, take the number five and its 2-neighbourhood. We are working out U(5, 2), which is the interval (5 − 2, 5 + 2) = (3, 7). The punctured neighbourhood would be R(5, 2) = (3, 5) ∪ (5, 7). It is the same neighbourhood, just without the point five. Here is the 2-neighbourhood of the point 5 drawn on a number line:

The punctured 2-neighbourhood would look the same, but without the blue point 5.

Accumulation point

Next we need the definition of an accumulation point of a set. An element a ∈ ℝ is an accumulation point of a set M ⊆ ℝ if every punctured neighbourhood of it, however small, contains some point of M. More precisely:

$$\mathrm{\forall }ϵ>0:R(a,ϵ)\cap M\ne \mathrm{\varnothing }$$

Examples: every real number is an accumulation point of the set of real numbers. If we choose the number 10, then in any punctured neighbourhood of this point we can always find a real number. For $ϵ=\frac{1}{100}$ we get the punctured neighbourhood (9.99, 10) ∪ (10, 10.01). Both intervals contain some real number – in fact they contain infinitely many real numbers. So however small an ϵ we take, we always get intervals containing infinitely many real numbers.

If we took the set of positive integers as M, we would not find any accumulation point. For example, the number $\frac{1}{2}$ is not an accumulation point, because for $ϵ=\frac{1}{10}$ we get the intervals (0.4, 0.5) ∪ (0.5, 0.6), and neither of them contains a positive integer.

Motivation for limits

We start with the easiest definition – a finite limit at a finite point. First let’s see what we actually want to work out. Look at the following graph of the sign function:

Take the point x₁ = 3. The value of the function at this point is 1, so sgn(3) = 1. The function also has the value 1 at all “nearby” points, for example sgn(2.5) = 1 and sgn(3.5) = 1. If we approach the point x₁ = 3 from the left or from the right, the function values are always equal to one.

But if we take the point x₂ = 0, then sgn(0) = 0, yet all the surrounding points have a different function value! For example $\text{sgn}(-\frac{1}{2})=-1$ and $\text{sgn}(\frac{1}{2})=1$. The function values from the right get close to 1, while the function values from the left get close to −1. And at the point x₂ = 0 itself we get the function value 0. Sneaky. The values that the function approaches near a point are called the limits of the function at that point.

Definition of a finite limit at a finite point

Let f be a function and let x₀ ∈ ℝ be an accumulation point of its domain dom(f). Then we say that L ∈ ℝ is the limit of the function f at the point x₀ if

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

Now let’s see what this definition means. A finite point here means that the point is a real number. Later we will talk about limits at points at infinity. Let’s try to work out this limit:

$$\underset{x\to 3}{lim}\frac{x}{3}=?.$$

For simple functions like this, the limit of the function at the point x₀ is equal to the function value at that point, f(x₀). Later we will define exactly what a “simple function” is. Now let’s draw everything in a picture:

The black line is the graph of the function $f(x)=\frac{x}{3}$, the red point x₀ is the point where we are looking for the limit, and the green point L is the resulting limit. Notice that we show the limit on the y-axis.

Let’s follow the definition. At the beginning of the definition we have (∀ϵ > 0), and ϵ is used later in |f(x) − L|<ϵ – this is the mathematical way of saying that we want all the values f(x) to be less than ϵ away from L. So we are looking for the ϵ-neighbourhood of the point L on the y-axis. According to the definition we can choose any ϵ > 0; let’s choose $ϵ=\frac{3}{10}$. We find the points on the y-axis that are exactly ϵ away from the point L:

All the points on the y-axis between these green lines are in the ϵ-neighbourhood of the point L. These are the points the definition talks about.

The definition continues with (∃δ>0), which is then used in the expression 0<|x − x₀|<δ. We are saying that the difference x − x₀ should be less than δ. Since x, x₀ lie on the x-axis, we will show the value of δ on the x-axis too. The expression 0<|x − x₀|<δ again tells us to take all the x that are less than δ away from x₀, but at the same time more than zero away, that is x ≠ x₀. This means that this expression describes the punctured δ-neighbourhood of the point x₀. So we can rewrite the definition equivalently using neighbourhoods of points:

$$(\mathrm{\forall }ϵ>0)(\mathrm{\exists }\delta >0)(\mathrm{\forall }x\in \mathrm{dom}(f))(x\in R(x_0,\delta )\Rightarrow f(x)\in U(L,ϵ))$$

Let’s go back to the previous definition and continue drawing the picture. To illustrate it, we need the next part of the definition:

$$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<ϵ$$

So for all x in the punctured δ-neighbourhood of the point x₀, some condition must hold. Let’s add the punctured δ-neighbourhood of the point x₀ to the picture, for example for $\delta =\frac{3}{2}$:

All the x between the red dashed lines belong to the δ-neighbourhood of x₀, except the point x₀ itself. Let’s look at the condition again:

$$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<ϵ$$

For all x in the punctured δ-neighbourhood of the point x₀, all the function values f(x) must lie in the ϵ-neighbourhood of the point L. So we take all the x from the punctured δ-neighbourhood of the point x₀ and work out the function values. In the next picture these function values are shown as a blue line segment:

The point x₀ is not in the punctured neighbourhood, which is why there is a hollow circle at that place. The definition tells us that all these blue points must lie in the ϵ-neighbourhood of the point L. In terms of the colours in the picture: all the blue points must lie between the green lines, which show the ϵ-neighbourhood. Is that true?

Of course not: for example, for $x=\frac{16}{10}$ we clearly get a blue point that is not between the green lines, that is, not in the ϵ-neighbourhood of the point L.

That doesn’t matter, though. The definition says that for every ϵ-neighbourhood of the point L there must exist some δ-neighbourhood such that … the rest of the condition holds. That means we only need one such δ-neighbourhood to exist. $\delta =\frac{3}{2}$ was clearly not a good choice, but we can try a different δ, for example $\delta =\frac{1}{2}$. The picture would change like this:

Again we check whether all the function values for all x in the punctured δ-neighbourhood of the point x₀ lie in the ϵ-neighbourhood of the point L. In short: are all the blue points between the green lines? Yes, now they are – hooray.

Does that mean that L is the limit of the function at the point x₀? Not necessarily, because the definition says (∀ϵ > 0) …. So far we have checked the definition for one particular ϵ, and there are still … infinitely many left.

In other words, to really show that $\underset{x\to 3}{lim}\frac{x}{3}=1$, we would have to find, for every ϵ-neighbourhood, some δ-neighbourhood for which the previous condition holds. From the picture we can see that we can always find such a δ-neighbourhood. For example, if we reduce ϵ to $ϵ=\frac{1}{10}$, then we can reduce δ to $\delta =\frac{2}{10}$ and it will work:

However small we make ϵ, we must always be able to find a δ such that all the function values end up in the ϵ-neighbourhood of the limit L.

Proof of a finite limit at a finite point

Let’s try to actually prove the previous limit. From the picture we already know that probably

$$\underset{x\to 3}{lim}\frac{x}{3}=1,$$

and now we will try to prove it mathematically. From the picture we might guess that if we always take δ to be the same value as ϵ, it could work. So let δ = ϵ. We now have to prove

$$L-ϵ<f(x)<L+ϵ,\text{for all}x\in (x_0-\delta ,x_0)\cup (x_0,x_0+\delta )$$

Since we set δ = ϵ, we change every δ to ϵ:

$$L-ϵ<f(x)<L+ϵ,\text{for all}x\in (x_0-ϵ,x_0)\cup (x_0,x_0+ϵ)$$

We substitute 1 for L and 3 for x₀:

$$1-ϵ<f(x)<1+ϵ,\text{for all}x\in (3-ϵ,3)\cup (3,3+ϵ)$$

Now we should prove all the inequalities. Since the function $\frac{x}{3}$ is increasing, it is enough to check the endpoints 3 − ϵ and 3 + ϵ. We start with 3 − ϵ – we substitute the value 3 − ϵ into the function $\frac{x}{3}$ and show that this value always satisfies the inequalities 1 − ϵ < f(x) < 1 + ϵ:

$$\begin{array}{rcl}1-ϵ<& f(x)& <1+ϵ\\ 1-ϵ<& \frac{x}{3}& <1+ϵ\\ 1-ϵ<& \frac{3-ϵ}{3}& <1+ϵ\\ 1-ϵ<& \frac{3}{3}-\frac{ϵ}{3}& <1+ϵ\\ 1-ϵ<& 1-\frac{ϵ}{3}& <1+ϵ\\ -ϵ<& -\frac{ϵ}{3}& <ϵ\end{array}$$

The same with 3 + ϵ:

$$\begin{array}{rcl}1-ϵ<& f(x)& <1+ϵ\\ 1-ϵ<& \frac{x}{3}& <1+ϵ\\ 1-ϵ<& \frac{3+ϵ}{3}& <1+ϵ\\ 1-ϵ<& \frac{3}{3}+\frac{ϵ}{3}& <1+ϵ\\ 1-ϵ<& 1+\frac{ϵ}{3}& <1+ϵ\\ -ϵ<& \frac{ϵ}{3}& <ϵ\end{array}$$

A wrongly chosen limit

Now let’s sketch what would happen if we thought that the limit of the function as x tends to three was one half. We would choose some small epsilon neighbourhood:

At this point we can no longer find a delta neighbourhood of the point x₀ whose function values would all fit between these green lines. We can see that for the chosen δ-neighbourhood all the blue points are outside the green lines. If we made the δ-neighbourhood bigger, we would only increase the number of blue points that we cannot get between the green lines.

The limit of the sign function

The sign function is a beautiful function that will probably tie your head in knots. Here is the graph:

The graph is not entirely clear, so here it is in words: if x is negative, then the sign function is minus one; if it is positive, it is plus one; and if x is zero, then the sign function is zero too. We modify the graph using the absolute value, f(x) = |sgn(x)|, and we get |sgn(0)| = 0 and |sgn(x)| = 1 in all other cases. The graph:

The question is, what is this limit equal to?

$$\underset{x\to 0}{lim}|\text{sgn}(x)|=?$$

The function is defined at zero and the function value is also zero. However, that doesn’t tell us that the limit will also be zero. A limit can differ from the function value at that point.

The only way to choose the limit so that the definition works is to choose the limit to be one. We can also use intuition: what does the function value approach as x approaches zero from the left? It keeps approaching one. It doesn’t bother us that it finally jumps to zero. The same goes for x approaching zero from the right.

$$\underset{x\to 0}{lim}|\text{sgn}(x)|=1$$

If we left the function in its original form, f(x) = sgn(x), then the function would not have a limit at zero; it would have two different one-sided limits. From the left the function approaches minus one and from the right plus one. You can try to prove this from the definition.

Limits that do not exist

Let’s look at a few more examples where limits do not exist.

The simplest example of a limit not existing is working out a limit at a point that is not an accumulation point of the domain of f. So, for example, this limit does not exist:

$$\underset{x\to -1}{lim}\ln x$$

The graph:

Periodic functions often have no limit at infinity (though not all of them!). Typical examples are trigonometric functions such as sine. The graph:

The limit at infinity simply does not exist: the value of sine keeps oscillating between one and minus one, so the limit cannot be worked out.