Limits
Let’s consider the following function defined over the set of real numbers $\displaystyle f(x) = \frac{e^{x}  1}{x}$. We can see that $f$ is undefined at $x = 0$ as there is no welldefined way of evaluating $f(0) = \frac{0}{0}$.
But, there is an interesting property we can note about $f$ near $x = 0$: when we take an arbitrary sequence of values that converge to $x = 0$ like $(1, 0.1, 0.01, 0.001,\dots)$, we can see that values mapped with $f$ approach a steady value.
\begin{align} f(1) &= 1.718\dots \\ f(0.1) &= 1.051\dots \\ f(0.01) &= 1.005\dots \\ f(0.001) &= 1.000\dots \end{align}
We continue to observe this behavior even if we were to ‘approach’ $x = 0$ from values below:
\begin{align} f(1) &= 0.632\dots \\ f(0.1) &= 0.951\dots \\ f(0.01) &= 0.995\dots \\ f(0.001) &= 0.999\dots \end{align}
In both cases, it appears that we can make $f$ as close to the value $1$ as we want just by continuing to evaluate $f$ over the sequence. This is in essence what it means for the limit of $f(x)$ to equal $1$ as $x$ approaches $0$. Symbolically, we can write this statement as
$$ \lim_{x\to 0} f(x) = 1. $$
Definition. $\lim_{x\to a} f(x) = L$ if for any sequence $(x_1, x_2, x_3,\dots)$ that converges to $x = a$, the sequence $(f(x_1), f(x_2), f(x_3), \dots)$ converges to $L$.
Remark: although this definition serves as a good starting point for understanding limits, there is an equivalent definition that is more commonly used in proofs: the epsilondelta definition.
Examples
1. $\displaystyle f(x) = \begin{cases} 0 &\text{if } x < 0 \\ 1 &\text{otherwise} \end{cases}$
The limit of $f$ at $x = 0$ does not exist as $f$ stays constant at 0 as we approach from below $x = 0$ and stays constant at $1$ as we approach from above $x = 0$. In this case, we can write
$$\lim_{x\to 0^{}} f(x) = 0,\quad \lim_{x\to 0^{+}} f(x) = 1.$$
Since $\lim_{x\to 0^{}} f(x) \neq \lim_{x\to 0^{+}} f(x)$, we say the limit $\lim_{x\to 0} f(x)$ does not exist.
2. $f(x) = \frac{1}{x^2}$
$f$ can get arbitrarily positively large as $x$ approaches $0$, so in these cases, we just say $f(x)$ diverges to positive infinity as $x$ approaches $0$. That is,
$$\lim_{x\to 0} f(x) = +\infty.$$
3. $f(x) = \frac{1}{x}$
In this case, $\lim_{x\to 0^{}} f(x) = \infty$ and $\lim_{x\to 0^{+}} f(x) = +\infty$, so we say the limit does not exist.
Properties

If $\lim_{x\to a} f(x) = L_1$ and $\lim_{x\to a} g(x) = L_2$ (i.e., these limits exist), then $\lim_{x\to a} f(x)g(x) = L_1L_2$.

If $\lim_{x\to a} f(x) = L_1$ and $\lim_{x\to a} g(x) = L_2$, then $\lim_{x\to a} [f(x) + g(x)] = L_1 + L_2$.

If $\lim_{x\to a} f(x) = L$ and $g(x)$ is continuous at $x = L$, then $\lim_{x\to a} g(f(x)) = g(L)$.