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 $$p$$.