limit function producing both an undefined and defined value when rewriting the polynomials of the function Here I have I have a perfect square trinomial:
$x^2 - 6x + 9$
Here I have a perfect square binomial:
$x^2 - 9$
I am asked to find the limit as x approaches 3 of $\frac{x^2 - 6x + 9}{x^2 - 9}$. The full equation is illustrated below:
$\lim \limits_{x \to 3} \frac{x^2 - 6x + 9}{x^2 - 9}$
If we plug 3 in for x, you'll notice the result is $\frac{0}{0}$. In other words, the value is undefined and that's typically how it is with these limit problems. We now need to find the values approaching 3 from both the negative and positive end. 
But this is where it gets weird for me. The polynomials can be simplified through factoring. 
$x^2 - 6x + 9$ can be factored into $(x - 3)(x - 3)$
$x^2 - 9$ can be factored into $(x + 3)(x - 3)$
Since there is a $(x - 3)$ in the numerator and denominator, we can cancel them out to get the following equation:
$\frac{x - 3}{x + 3}$
That means the limit can be rewritten as:
$\lim \limits_{x \to 3} \frac{x - 3}{x + 3}$
But now if we plug in 3, we get $\frac{0}{6}$ which is 0. 
How is this possible? How does rewriting the equation give a different answer? More importantly, how does simply rewriting the question make an undefined value suddenly defined?
 A: When you get 0/0, you cannot conclude that the limit is undefined.  0/0 is an "indeterminate form", which means it tells you nothing about the limit.  When you get 0/0, the limit could be defined or undefined. You must find a different approach, as you did in the second approach.
A: That's a great question. When you take a limit as $x\to 3$, you are concerned with all points around $x=3$, but not actually $x=3$. In any textbook, this is the standard meaning of a limit. By defining it in this way, it allows you to consider the behavior near points.
When you plugged in and got $0/0$, that's an indication that if the limit exists, the function has a hole there. You can always fill in the hole by defining $f(3)=L$, where $L$ is the limit.
Now why did it work out? It's because when you factored and got $\frac{(x-3)(x-3)}{(x-3)(x+3)}$, the reason you were able to cancel the $x-3$ terms is because of what I discussed above. For $x$ close to $3$, but not equal to it, we have $x-3 \neq 0$. Hence, you have a fraction with a common nonzero term on the numerator and denominator so you can cancel them.
When you take the limit of $\frac{x-3}{x+3}$, again you are concerned with all points near $x=3$. It turns out though, the function $f(x)=\frac{x-3}{x+3}$ is continuous at $x=3$ so you are really just observing that $\displaystyle f(3)=\lim_{x\to 3} f(x)=0$. There are no contradictions; you are always plugging in numbers close to $3$, and it's just a coincidence that this was allowed in the second case.
