$$\lim_{x\to{c}}f(x) = L$$
In words:
The limit of the function $f(x)$ as $x$ approaches $c$ is the value of the function $L$ at that point
Or from Wikipedia:
$f(x)$ can be made to be as close to $L$ as desired by making $x$ sufficiently close to $c$. In that case, the above equation can be read as "the limit of $f$ of $x$, as $x$ approaches $c$, is $L$
Or mathematically:
Let $\delta$ and $\epsilon$ be two positive numbers:
$$ \begin{aligned} \delta &\in \mathbb{R}_{>0}\\ \epsilon &\in \mathbb{R}_{>0} \end{aligned} $$
Then we can write:
$$\lim_{x\to{c}}f(x) = L$$
Means that for all $x \ne c$
$$\lvert x−c \lvert < \delta \implies \lvert f(x)−L\lvert < \epsilon$$
So for every $\epsilon$ there is a $\delta$ such that the above formula holds.
Let us define following piecewise function
$$f(x) = \begin{cases} \frac{x^2}{100}+10 & \text{if } x < 0\\ 20 & \text{if } x = 0\\ \frac{x^2}{100}+10 & \text{if } x > 0\\ \end{cases} $$
The limit $L$ of the piecewise function for $x\to{0}$ is $10$
You can use the following slider to regulate $\epsilon$
You can use the following slider to regulate $\delta$
Watch how no matter how small you make $\epsilon$, you can always find a $\delta$ such that the $\epsilon$-$\delta$ definition still holds. If the definiton does not hold, the vectors representing $\delta$ willbe coloured red
Let us define following piecewise function
$$f(x) = \begin{cases} \frac{x^2}{100}+5 & \text{if } x < 0\\ 10 & \text{if } x = 0\\ \frac{x^2}{100}+15 & \text{if } x > 0\\ \end{cases} $$
The limit $L$ of the piecewise function for $x\to{0}$ comming from the left is $5$. However, the right-sided limit is 15. Which means the above delta-epsilon definition of limit does not hold!
Let us see what we get when we suppose the limit is 5:
You can use the following slider to regulate $\epsilon$
You can use the following slider to regulate $\delta$
Everything from the above still holds, until we let $\epsilon$ get smaller then 10. From that moment there is no delta for the right side of the piecewise fucntion as there is no value defined for $x$ which result in the desired value of $y=f(x)$