I have this rational expression to evaluate, $$ {{3a-3}\over {4a(a-1)}} \text { if } a=1. $$
I understand that if you substitute 1, both the numerator and denominator would turn out 0, thus making it indeterminate. But what if I factor out the numerator $3a-3$ to $3(a-1)$ and cancel out $(a-1)$ from both the numerator and denominator, wouldn't it just give $3\over4$ as the answer?
For this function:
$$\frac{3a-3}{4a(a-1)}$$
When $a=1$ this function will not be defined at that specific x-value.
The reason is because:
$$\frac{3a-3}{4a(a-1)} = \frac{3}{4a}$$
Even though the $(a-1)$ term cancels, the function still isn't defined at that point. We call $a = 1$ to be a hole in the function.
You can read more about it here
Comment if you have any questions.