Limit of a function

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 neighbourhood of a point

We begin by defining the epsilon neighborhood of a point on the set of real numbers. Let r ∈ ℝ. Then we say that the epsilon neighborhood of this point for ε > 0 is the open interval (r−ε, r+ε). The reduced epsilon neighborhood is the same, except that it does not contain the point r. The reduced neighborhood is (r−ε, r) ∪ (r, r+ε). We denote the epsilon neighborhood of the point a by U(a, ε), and the reduced neighborhood by R(a, ε).

For an example, consider the number five and its 2-neighborhood. Thus we compute U(5, 2), which is the interval (5 − 2, 5 + 2) = (3, 7) The reduced neighborhood would be R(5, 2) = (3, 5) ∪ (5, 7). The same neighborhood, only without the point five. Graphically represented by 2-the neighborhood of the point 5:

The reduced 2-neighborhood would look the same, except that the blue point 5 would not be included.

Bulk point

Next, we need to know the definition of a bulk point of a set. An element a ∈ ℝ is a bulk point of a set M ⊆ ℝ, if it holds that in every reduced neighborhood of it, however small, there is a point of the set M. More precisely:

$$\forall \epsilon > 0: R(a, \epsilon) \cap M \ne \emptyset$$

Examples: every real number is a mass point of a set of real numbers. If we choose the number 10, we can always find a real number in any reduced neighborhood of this point. For $\epsilon=\frac{1}{100}$ we get the reduced neighborhood of (9,99; 10) ∪ (10; 10,01). In both intervals there is a real number - more precisely there are infinitely many real numbers. So no matter how small we take ε, we always get intervals that contain infinitely many real numbers.

If we were to choose the set of natural numbers as M, we would find no bulk point. For example, the number $\frac12$ is not a mass point because for $\epsilon=\frac{1}{10}$ we get the intervals (0,4; 0,5) ∪ (0,5; 0,6), neither of which contains any natural number.

Motivation for limits

We start with the lightest definition - an eigenlimit at an eigenpoint. First, let us see what we actually want to compute. Look at the following graph of the signum function:

If we take the point x₁ = 3, the function value at this point is 1. It is also true that $\mbox{sgn}(3)=1$. Yet for all "nearby" neighboring points, the function has the function value 1, for example $\mbox{sgn}(2,5)=1$ or $\mbox{sgn}(3,5)$. If we approach the point x₁ = 3 from the left or right, the function values are always equal to one.

But if we take the point x₂ = 0, then $\mbox{sgn}(0) = 0$ holds, while all the surrounding points have a different functional value! For example, $\mbox{sgn}(-\frac12)=-1$ or $\mbox{sgn}(\frac12)=1$. The function values from the right approach the value 1, while the function values from the left approach the value −1. Finally, at the point x₂ = 0 we get the function value 0. Betrayal. These values that the function approaches at the point are then called the limits of the function at the point.

Definition of eigenlimits at a point

Let f be a function, x₀ ∈ ℝ be a bulk point of the definition domain of the function f. Then we say that L ∈ ℝ is the limit of the function f at the point x₀, if

$$(\forall \epsilon > 0),(\exists\delta>0),(\forall x\in D(f)),(0<|x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon).$$

Now what this definition means. A proper point in this case means that the point is a real number. Later we will discuss limits at non-eigenpoints, which will mean infinity. Let's try to calculate this limit:

$$ \lim_{x\rightarrow3} \frac{x}{3} = ?. $$

For functions this simple, the limit of the function at a point x₀ is equal to the function value at that point: f(x₀). Later, we will define exactly what "simple function" means. Now let's plot everything in the figure:

The black line represents the graph of the function $f(x)=\frac{x}{3}$, the red point x₀ represents the point at which we are looking for the limit, and the green point L represents the resulting limit. Note that we plot the limit on the axis y.

We proceed as defined below. At the first point of the definition, $(\forall \epsilon > 0)$ is and later ε is used in |f(x)−L|<ε - this is a mathematical notation such that we want all f(x) to be less than ε away from L. Thus we are looking for ε-the neighborhood of the point L on the axis y. According to the definition, we can choose any ε > 0, we choose for example $\epsilon = \frac{3}{10}$. We find the points on the axis y that are just ε away from the point L:

All the points on the y axis that are between these green lines are located in ε-around the point L. These are the points that the definition is talking about.

The definition continues with the expression $(\exists\delta>0)$, which is further used in the expression 0<|x − x₀|<δ. We are saying that some difference x − x₀ should be less than δ. Since x, x₀ is on the x axis , we will also represent the value of δ on the x axis. In doing so, the expression 0<|x − x₀|<δ again tells us to take all x, which are less than δ, but also more than zero, from x₀, i.e. x ≠ x₀. This means that this notation gives us the reduced δ-environment of the point x₀. Thus we can equivalently rewrite the definition of the limit using the neighborhood of points:

$$(\forall \epsilon > 0),(\exists\delta>0),(\forall x\in D(f)),(x\in R(x_0, \delta) \Rightarrow f(x) \in U(L, \epsilon))$$

We return to the previous definition and continue drawing the figure: To illustrate this, we need the next part of the definition:

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

Thus, for all x of the reduced δ neighborhood of the point x₀, the following must hold... some condition. Let's add the reduced δ-environment of the point x₀, for example for $\delta=\frac{3}{2}$, to the picture:

All x, which is between the red dashed lines, belongs to the δ-environment x₀, except for the point x₀ itself. Again, we look at the previous condition:

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

For all x from the reduced δ -environment of the point x₀, it must hold that all functional values of f(x) are in the ε-environment of the point L. So we take all x from the reduced δ-environment of the point x₀ and compute the functional values. These functional values are represented by the blue line in the following figure:

The point x₀ is not in the reduced neighborhood, hence the circle at that point. The definition tells us that it must be true for all these blue points that they are in the ε-environment of the point L. To put it in the colors of the figure: all the blue points must be located between the green lines that represent the ε-environment. Is this valid?

Of course it doesn't, for example for $x=\frac{16}{10}$ we obviously get some blue point that is not between the green lines, i.e. not in the ε-environment of L.

But that doesn't matter, the definition tells us that for every ε neighborhood of the point L there must be some δ-neighborhood such that ... the rest of the condition. That is, it is enough for us if there is one such δ-environment. $\delta=\frac{3}{2}$ Obviously this was not a good choice, but we could try another δ, for example $\delta=\frac12$. The picture would change like this:

Again, we look to see if all the function values for all the x of the reduced δ-environment point x₀ are in the ε-environment point L. In short, if all the blue points are between the red lines. Yes, they are now, hooray.

Does this mean that L is the limit of the function at point x₀? Not necessarily, because the definition says that $(\forall \epsilon > 0) \ldots$. We have so far checked the definition for one particular ε, there are still ... infinitely many more to go.

In other words - for $\lim_{x\rightarrow3} \frac{x}{3} = 1$ to really hold, we would have to find for every ε-environment some δ-environment such that the previous condition holds. From the figure we can see that we can always find such a δ-environment, for example if we reduce ε to $\epsilon=\frac{1}{10}$, then we can reduce δ to $\delta=\frac{2}{10}$ and it will work:

No matter how small we take ε, we always have to find δ such that we can get the function values into the ε-environment of the limits of the function L.

Proof of the eigenlimit at the eigenpoint

Let's try to actually prove the previous limit. We already know from the figure that we probably

$$ \lim_{x\rightarrow3} \frac{x}{3} = 1, $$

Now we'll try to prove it mathematically. From the figure we could deduce that if we always take the same value for δ as ε, it might work. So let's put δ = ε. Now we have to prove

$$ L - \epsilon < f(x) < L + \epsilon, \quad \mbox{ For all }\quad x \in (x_0-\delta, \delta)\cup(x_0, x_0+\delta) $$

Since we put δ = ε, we change all δ to ε:

$$ L - \epsilon < f(x) < L + \epsilon, \quad \mbox{ For all }\quad x \in (x_0-\epsilon, \epsilon)\cup(x_0, x_0+\epsilon) $$

After L we put 1, and after x₀ we put 3:

$$ 1 - \epsilon < f(x) < 1 + \epsilon, \quad \mbox{ For all }\quad x \in (3-\epsilon, \epsilon)\cup(3, 3+\epsilon) $$

Now we should prove all the inequalities. Since the function $\frac{x}{3}$ is increasing, we just need to verify the extreme points 3−ε and 3+ε. Starting with 3−ε - we will insert the value ε − 3 into the function $\frac{x}{3}$ and show that this value will always satisfy the inequality 1 − ε < f(x) < 1 + ε:

\begin{eqnarray} 1 - \epsilon < &f(x)& < 1 + \epsilon 1 - \epsilon < &\frac{x}{3}& < 1 + \epsilon 1 - \epsilon < &\frac{3-\epsilon}{3}& < 1 + \epsilon 1 - \epsilon < &\frac{3}{3}-\frac{\epsilon}{3}& < 1 + \epsilon\ 1 - \epsilon < &1-\frac{\epsilon}{3}& < 1 + \epsilon\ -\epsilon\ &-\frac{\epsilon}{3}& < \epsilon\ \end{eqnarray}

Same with 3+ε:

\begin{eqnarray} 1 - \epsilon < &f(x)& < 1 + \epsilon\ 1 - \epsilon < &\frac{x}{3}& < 1 + \epsilon\ 1 - \epsilon < &\frac{3+\epsilon}{3}& < 1 + \epsilon\ 1 - \epsilon < &\frac{3}{3}+\frac{\epsilon}{3}& < 1 + \epsilon\ 1 - \epsilon < &1+\frac{\epsilon}{3}& < 1 + \epsilon\ -\epsilon\ < &\frac{\epsilon}{3}& < \epsilon\ \end{eqnarray}

Wrong limit

Now let's sketch what would happen if we thought the limit of the function for x approaching three was one half. We would choose some small neighborhood of epsilon:

And at this point we can no longer find such a delta neighborhood of the point x₀, whose function values would all fit between these green lines. We see that for the chosen δ neighborhood, all the blue points are outside the green lines. If we enlarge the δ-neighborhood, we only increase the number of blue points that do not fit between the green lines.

Limita signum

Signum is a nice feature that will probably mess with your head. The graph follows:

The graph is not entirely clear, so a verbal description: if x is negative, then signum is minus one, if it is positive, then it is plus one, and if x is zero, then signum is zero as well. We modify the graph using the absolute value of f(x) = |sgn(x)| to get sgn(0) = 0, for the other cases sgn(x) = 1. Graph:

The question is, what is this limit equal to?

$$\lim_{x\rightarrow0}|\mbox{sgn}(x)|=?$$

The function is defined at point zero and the function value is also equal to zero. However, this does not tell us that the limit will also be equal to zero. The limit may be different from the function value at that point.

The only way to choose the limit so that we get the definition right is to choose the limit at one. We can also use intuition to help us: what is the functional value if x approaches zero from the left? It is constantly approaching one. That it eventually jumps to zero doesn't bother us. Similarly for x approaching zero from the right.

$$\lim_{x\rightarrow0}|\mbox{sgn}(x)|=1$$

If we keep the function in its original form, i.e. f(x) = sgn(x), then the function at zero will have no limit, it will have two different one-sided limits. From the left, x will approach minus one and from the right it will approach plus one. You can try to prove this by definition.

Non-existent limits

We'll show a few more examples where limits don't exist.

The simplest example of a non-existent limit is to compute the limit at a point that is not the bulk point of D(f). So, for example, there is no limit:

$$\lim_{x\rightarrow-1}\ln x$$

Graph:

Periodic functions often have no limit at infinity (not true for all!). Typically goniometric functions like sine. The graph follows:

The limit at infinity simply does not exist, the value of sin constantly oscillates between one and minus one, so the limit cannot be calculated.

External links and resources

• Jan Tomeček, Mathematical Analysis 1
• Solved examples at CUNI [PDF]