Limit of a function in a point

Considering a function $f(x)$ we say that it has limit $L$ in a point $p$ if $f(x)$ takes values as closed to $L$ as we want, taking sufficiently close points to $p$ but different from $p$. This concept is denoted as:

$$\lim_{x \to p}{f(x)}=L$$

Let's take the function $f(x)=\dfrac{x^2-1}{x}$. If we look for the limit of the function at point $x=5$, we will see:

$$\lim_{x \to 5}{f(x)}=\lim_{x \to 5}{\dfrac{x^2-1}{x}}=\dfrac{5^2-1}{5}=\dfrac{24}{5}$$

In this case, the function $f(x)$ coincides with its limit at point $x=5$.

It can seem that the function always coincides with its limit at any point, but this is not the case. In the following example we see a case in which the function does not coincide with its limit:

$$f(x)=\left\{\begin{array}{c} 1 \ \text{ si } x\neq0 \\ 3 \ \text{ si } x=0 \end{array} \right.$$

In this case we see that $f(0)=3$, but:

$$\lim_{x \to 0}{f(x)}=\lim_{x \to 0}{1}=1$$

This means that close to $x=0$ the function always takes value $1$ and, consequently, its limit is $1$. However, the function at $x$ equals zero has value $3$.

This example is a clear example of discontinuous function. The discontinuous functions are detected easily since in the discontinuity points, the limits and the function do not coincide.

Therefore, to make the limit of a function $f(x)$ at point $p$ involves seeing all the values of the function $f(x)$ when we are located very close to $x=p$, but not exactly on $p$.