A limit of a function

Limit is a basic, but nonetheless fundamental instrument for studying the behavior of a function. Definition of limit is not only encountered when we talk about functions, but also when we observe and study sequences. We use the notion of limits in order to approximate derivatives at given points, and this makes limits so useful, moreover, we can see that limits find many applications in different disciplines ranging from engineering to see how a function changes or chemistry to see how different processes are carried out.

What is a limit of a function?

In general, limits allow us to determine what value a function is approaching when we use a particular input. Recall the pendulum oscillations? With time, any pendulum stops, so we can say that as time increases, the amplitude of oscillation decreases. And as the time approaches 0, our pendulum approaches its initial position. It seems logical, but physicists can prove it mathematically with limits.

We define it this way:

$\lim\limits_{x \to a} f(x) = A$

Can we just put $a$ instead of $x$ in the function? Yes, if our function has no undefined points in $a$, in other words, if it is continuous.

$$\lim\limits_{x \to a} f(x) = f(a), \text{if } f \ \text{is a continuous function at$$$a$}.There are numerous functions where we can't plug in a number to find the limit. So, the limits are used to describe what value a function is approaching as $x$ approaches $a$, but it is possible that the function can be undefined at $a$.

Let's consider the following example. Suppose we have a function $g(x) = \dfrac{x^2 +x - 2}{x-1}$ and we want to know the value this function approaches when $x$ gets closer and closer to $1$:

$$\lim\limits_{x \to 1} \dfrac{x^2 +x - 2}{x-1} .$$

If we try to substitute the value $x=1$ into this function, we get

$$f(1) = \dfrac{1^2+1-2}{1-1} = \dfrac{0}{0} \ne 3.$$

So we see that our function is not defined at $1$.

Let's put some inputs like $0.9, \ 0.99, \ 0.999, \ldots$ into it:

As you can see, for a sequence of arguments ${0.9, 0.99, 0.999, ...}$ function approaches $3$. It can actually be proven that for any sequence of arguments $x_n$, which approaches $1$, but does not reach $1$ (i.e., limit of sequence $\lim \limits_{n \to \infty} x_n = 1$ and $x_n \neq 1$ for any $n$), the function value will approach $3$ (i.e. limit of sequence $\lim\limits_{n \to \infty} f(x_n) = 3$). By the way, this is the definition of the limit of a function according to Heine!

So,

$$\lim\limits_{x \to 1} \dfrac{x^2 +x - 2}{x-1} = 3.$$Or we can consider $\dfrac{x^2 +x - 2}{x-1}$ as $\dfrac{(x+2)(x-1)}{x-1}$and divide by $x-1$. Here we should take into account that $x$ only approaches to $1$, but never is equal to it. We will get

$\lim\limits_{x \to 1} \dfrac{x^2 +x - 2}{x-1} = \lim\limits_{x \to 1} \dfrac{(x+2)(x-1)}{x-1}= \lim\limits_{x \to 1} x+2 = 3$

These examples show how limits emphasize the behavior of a function and what they get arbitrarily close to.

But what does "$f(x)$ is arbitrarily close to a number" mean? Let's look at the formal definition of a limit of a function.

The formal definition of a limit

A limit of a function $f(x)$ as $x$ approaches $a$ is the number $A$ if the following condition is met:

$$\text{if for every } \varepsilon>0, \text{there exists } \delta>0, \\ \text{such that for all } x \ \text{we have } \\ 0< |x-a| < \delta \ \Rightarrow \ |f(x)- A| < \varepsilon.$$

A constant number $A$ is called the limit of the function $f(x)$ as $x$ tends to $a$.

$\varepsilon$ is a given positive arbitrary small number, and we can find some positive $\delta$ that is dependent on it. We must follow this condition: for all $x$ from $\delta$-neighbourhood of $a$ and $0< |x-a| < \delta$, all $f(x)$ will be in $\varepsilon$-neighbourhood of $A$.$$\lim\limits_{x \to a} f(x) = A.$$This definition means that when $x$ gets close to $a$, then $f(x)$ gets close to $A$. So, if $|x-a|$ is small, then $|f(x) - A|$ is also small.

Let's prove that $\lim\limits_{x \to 1} \dfrac{x^2 +x - 2}{x - 1} = 3$ using the formal definition.

$|x-1|<\delta \ \Rightarrow \ | \dfrac{x^2 +x - 2}{x - 1} - 3|<\varepsilon.$

We need to find $\delta$, so let's solve the inequality:

$| \dfrac{x^2 +x - 2}{x - 1} - 3|<\varepsilon,$

$-\varepsilon< \dfrac{x^2 +x - 2}{x - 1} - 3<\varepsilon,$

$-\varepsilon< \dfrac{(x+2)(x-1)}{x - 1}-3 <\varepsilon,$

$-\varepsilon< (x+2) -3 <\varepsilon, \\ -\varepsilon< x-1 <\varepsilon$

Remember that we are trying to build $|x-1|<\delta$. We make a module instead of the inequality:

$-\varepsilon< x-1 <\varepsilon, \\|x-1| < \underbrace{\varepsilon}_{=\delta}.$

That's how we found our delta: $\delta = \varepsilon$. So, for every $\varepsilon>0$ there is a delta $\delta = \varepsilon >0$ such that if

$|x-1|<\delta \ \Rightarrow \ | \dfrac{x^2 +x - 2}{x - 1} - 3|<\varepsilon.$

Then $\lim\limits_{x \to 1} \dfrac{x^2 +x - 2}{x - 1} = 3$

What if x approaches infinity?

We may want to know the end behavior of our function when $x$ approaches infinity in either direction: $x \to + \infty$ or $x \to - \infty$. In this case, the role of limits is the most noticeable: it is not always possible to determine by the graphic of a function how it behaves at large values of $x$, and the limits help us to figure this out.

For example, let function $\varphi(t)$ describe an angle that the pendulum swings away from vertical. Here $t$ stands for time. It is quite logical that any pendulum ceases to oscillate over time. In the language of mathematics, it could be written as $\lim\limits_{t \to +\infty} \varphi(t) = 0\text{.}$

The formal definition of such limits will be:

$$\text{if for every } \varepsilon>0,\ \text{there exists } \delta>0, \\ \text{such that for all } |x|>\delta, |f(x)- A| < \varepsilon\\\lim\limits_{x \to \infty } f(x) = A.$$

$$\text{if for every } \varepsilon>0, \ \text{there exists } \delta>0, \\ \text{such that for all } x>\delta, |f(x)- A| < \varepsilon\\\lim\limits_{x \to +\infty } f(x) = A.$$

$$\text{if for every } \varepsilon>0, \ \text{there exists } \delta>0, \\ \text{such that for all } x<-\delta ,|f(x)- A| < \varepsilon\\\lim\limits_{x \to -\infty } f(x) = A.$$

We see that get rid of the module in the definition of limit $x \rightarrow +\infty$ because $x$ is obviously positive, and vice versa for limit $x \rightarrow -\infty$.

Also, it's important to know that $\dfrac{1}{0}$ approaches to $\infty$ because it has an infinitely small denominator, which makes the whole fraction infinitely big in the module. But we will study it in greater detail in our next topics.

Conclusion

In this topic, we have learned about the concept of a limit of a function. Let's now list all the notable points that we have seen:

• Limits help us describe a function or a sequence, to be precise, limits show which value this function or sequence is approaching depending on the input.
• If a function is continuous, then the following holds true: $\lim\limits_{x \to a} f(x) = f(a)$, for all values of $x$ and $a$.
• If a function is approaching a limit, it will get closer and closer to the value, but it will never reach it.
• If for every $\varepsilon>0$, there exists $\delta>0$, s.t. for all $x$ the following holds: $0< |x-a| < \delta \ \Rightarrow \ |f(x)- A| < \varepsilon$. $A$ will be the limit of $f(x)$ as $x$ tends to $a$.