Removable discontinuity is a specific type of discontinuity in a function where the function is undefined at a particular point, but can be made continuous by defining or redefining the function at that single point. This often occurs when a factor in the numerator and denominator of a rational function cancels out.

A function $f(x)$ has a removable discontinuity at $x=c$ if the following conditions are met:

1. $f(c)$ is undefined.
2. The limit of $f(x)$ as $x$ approaches $c$ exists (i.e., $\lim_{x\to c} f(x) = L$ for some finite number $L$).
3. If we were to define $f(c) = L$, the function would become continuous at $x=c$.

The most common way to spot a removable discontinuity is by simplifying a rational function. If a factor $(x-c)$ can be canceled from both the numerator and the denominator, then there is a removable discontinuity at $x=c$.

Let's analyze the function $g(x) = \frac{x^2 - 9}{x - 3}$.

1. Identify potential discontinuity: The denominator is zero when $x-3=0$, so $x=3$. Thus, $g(3)$ is undefined.
2. Evaluate the limit: $\lim_{x\to 3} \frac{x^2 - 9}{x - 3} = \lim_{x\to 3} \frac{(x-3)(x+3)}{x - 3}$.
3. Simplify: Since $x \to 3$, $x \neq 3$, we can cancel $(x-3)$: $\lim_{x\to 3} (x+3) = 3+3 = 6$.
4. Conclusion: The limit exists and is 6, but $g(3)$ is undefined. Therefore, $g(x)$ has a removable discontinuity at $x=3$. We can define a new function, $G(x)$, that is continuous at $x=3$: $G(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \ 6 & \text{if } x = 3 \end{cases}$ This function $G(x)$ is equivalent to $x+3$ for all $x$, and it is continuous everywhere.

It's important to distinguish removable discontinuities from other types, such as jump discontinuities (where the left and right limits exist but are different) and infinite discontinuities (where the limit approaches infinity, often due to vertical asymptotes). For removable discontinuities, the key is that the limit exists.