Suppose $D$ is a set of real numbers,  $f : D \rightarrow R$ with $a \in D$.  For this "one-sided" treatment we do not assume $D$ is an open interval.

Definition: We say that $L$ is the limit of $f$ as $x$ approaches $a$ from above, that is,  $$\lim_{x \rightarrow a^+} f(x) = L$$  if the following condition is satisfied:
For any $\epsilon \gt 0$, there exists a positive number $\delta (\ \gt 0)$ where if $x \in D$ and $0 < x - a < \delta$ then $|f(x) - L| < \epsilon$.
We can break down the analytic definition for a number $L$ to be a limit for $f$ as $x$ approaches $a$ from above into three steps needed to verify the definition if being correctly applied.
1. "For any $\epsilon \gt 0$": First consider a positive real number, $\epsilon$, that is arbitrarily chosen and usually considered to be small, that is "close to $0$".
The number $\epsilon$ will be a control on the values of the function $f$, ensuring that these values are "close to $L$".
In fact, we want $|f(x) - L| < \epsilon$; that is, we want the values of $f$, $f(x)$, to be within the interval $(L- \epsilon, L + \epsilon)$.
2. "there exists a positive number $\delta (\ \gt 0)$": Now based on the chosen $\epsilon$, the task is to find a second  positive number, $\delta \gt 0$. The number $\delta$ provides a control on the number $x$ in the domain of $f$ to keep $x$ "close to and greater than $a$".
In fact the constraint $0 < x - a < \delta$ amounts to having $x$ in the interval $(a,a+\delta)$.
3. "if $x \in D$ and $x - a < \delta$ then $|f(x) - L| < \epsilon$": The number $\delta$ works in the definition for the given $\epsilon$ only if when $x$ is a number with $x \in D$  and $0 < x - a < \delta$ then it can be demonstrated that $|f(x) - L| < \epsilon$. The demonstration will connect the constraint on $x$ with the actual function's definition and an analysis that eventually results in a conclusion connecting to the inequality $|f(x) - L| < \epsilon$.

##### CCD.LC.LimF1SVVisualization of one sided limit (from above) definition

Definition: We say that $L$ is the limit of $f$ as $x$ approaches $a$ from below, that is,  $$\lim_{x \rightarrow a^-} f(x) = L$$  if the following condition is satisfied:
For any $\epsilon \gt 0$, there exists a positive number $\delta (\ \gt 0)$ where if $x \in D$ and $0 < a - x < \delta$ then $|f(x) - L| < \epsilon$.

We can break down the analytic definition for a number $L$ to be a limit for $f$ as $x$ approaches $a$ from below into three steps needed to verify the definition if being correctly applied.
1. "For any $\epsilon \gt 0$": First consider a positive real number, $\epsilon$, that is arbitrarily chosen and usually considered to be small, that is "close to $0$".
The number $\epsilon$ will be a control on the values of the function $f$, ensuring that these values are "close to $L$".
In fact, we want $|f(x) - L| < \epsilon$; that is, we want the values of $f$, $f(x)$, to be within the interval $(L- \epsilon, L + \epsilon)$.
2. "there exists a positive number $\delta (\ \gt 0)$": Now based on the chosen $\epsilon$, the task is to find a second  positive number, $\delta \gt 0$. The number $\delta$ provides a control on the number $x$ in the domain of $f$ to keep $x$ "close to and greater than $a$".
In fact the constraint $0 < a -x < \delta$ amounts to having $x$ in the interval $(a-\delta,a)$.
3. "if $x \in D$ and $a - x < \delta$ then $|f(x) - L| < \epsilon$": The number $\delta$ works in the definition for the given $\epsilon$ only if when $x$ is a number with $x \in D$  and $0 < a - x < \delta$ then it can be demonstrated that $|f(x) - L| < \epsilon$. The demonstration will connect the constraint on $x$ with the actual function's definition and an analysis that eventually results in a conclusion connecting to the inequality $|f(x) - L| < \epsilon$.