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An indeterminate form occurs when evaluating a function results in an ambiguous expression like \( \frac{0}{0} \). This often happens when an expression has variables both in the numerator and the denominator that simultaneously become zero. In our function \( f(x) = \frac{(x+2)(x^2 - 1)}{(x+2)(x-1)} \), the denominator becomes zero at \( x = -2 \) and \( x = 1 \).
These are the points where we might suspect an indeterminate form because the whole expression seems to 'break down'. Not every zero in the denominator results in an indeterminate form, but if the numerator is zero at these points too, we are dealing with \( \frac{0}{0} \).
To resolve this, we look to simplify the function, hoping to remove the common factors that lead to the indeterminate form.

Simplifying rational functions is like decluttering. You want to remove all unnecessary parts that add complications. For the function \( f(x) = \frac{(x+2)(x^2 - 1)}{(x+2)(x-1)} \), we first need to simplify the numerator \((x+2)(x^2 - 1)\).
The expression \( x^2 - 1 \) can be rewritten using the difference of squares formula \( (x-1)(x+1) \). Thus, the numerator becomes \((x+2)(x-1)(x+1)\).
Now, both the numerator and denominator share the common factors \( (x+2) \) and \( (x-1) \), which can be canceled, leading to the simplified expression \( x+1 \). This process is essential as it helps in determining where the function is actually defined, apart from removing superfluous complications.

Discontinuities occur where a function is not well-defined or breaks unexpectedly. In our example, simplifying the function yielded \( x+1 \), but it isn't valid everywhere.
The initial function had zeros both in the numerator and denominator at \( x = -2 \) and \( x = 1 \), leading to the \( \frac{0}{0} \) indeterminate form. These points are where the holes (discontinuities) exist.
Though the simplified expression \( x+1 \) seems continuous everywhere, it is crucial to note that the original expression is undefined at \( x = -2 \) and \( x = 1 \). By acknowledging these points as holes, the modified function \( g(x) = x+1 \) is defined without them, effectively covering the same graph without breaks.