$$\lim_{x\to4} \frac{x^2-16}{x-4}$$
Is it correct to say that this limit is not defined because it has a point discontinuity at x=4, or to say that it does not exist because because of the point discontinuity? Or does it exist but has a point discontinuity?
Also, my teacher said that it can be factored so that it becomes $\lim_{x\to4}x+4$. In this case it would have a limit of 8.
So which of these answers is correct for $\lim_{x\to4} \frac{x^2-16}{x-4}$?
Your teacher is correct. The function does not have to be defined at $4$ to have a limit at $4$. It has to be defined on $(4-\epsilon,4)\cup(4,4+\epsilon)$ for some positive $\epsilon$ and it is. On that set it equals $x+4$.
Recall, again, that sometimes we cannot just plug in a value $a$ into a function as $x\to a$. This is one such case. We are not looking for the function value at $x=4$, where clearly, the function is undefined. We are looking for the value that the function approaches as $x$ gets very, very close to $4$.
Note $$\frac{x^2-16}{x-4} = \frac {(x-4)(x+4)}{x-4} \overset{x\neq 4}{=} x+4$$
Now, taking the limit $$\lim_{x\to 4} \frac{x^2-16}{x-4}=\lim_{x\to 4} x+ 4 = 8$$
Note that $x = 4$ is a removable discontinuity. The functions $x+ 4$ and $\dfrac{x^2 -16}{x-4}$ are equivalent at all values of $x$ except for x=4. But that doesn't matter here, since the limit (behavior) of the function as $x\to 4$ from the right and from the left agrees. The fact is, there is a single hole in the original graph at $x = 4$, but the function behaves, everywhere else, as does the line $f(x) = x+4$.
Your teacher is correct.
$$\require{cancel}\lim_{x\to4}\frac{x^2-16}{x-4}=\lim_{x\to4}\frac{\cancel{(x-4)}(x+4)}{\cancel{(x-4)}}=\lim_{x\to4}\,x+4=8$$
It does have a discontinuity at $x=4$, but you are not examining the behaviour for $x\color{red}=4$, but for $x$ that approaches $4$. In this case, you can cancel terms, and simplify your limit.
The limit of a function at a point does not have to equal the value of the function at that point, even if the function is defined there. We can say the function is continuous at a point if the limit at the point is the same as the value at the point.
With your function $f(x)=\frac {x^2-16}{x-4}$ you may encounter such a function in various circumstances.
For example, you may have been told that $f(x)$ is a continuous function, and have achieved this expression for $f(x)$ by some algebraic manipulation. Then you need to know the limit of $f(x)$ at $x=4$, because that is the point at which your expression doesn't work. Choosing the right value preserves continuity.
Or someone may say that $f(x)$ is given by this expression, except that $f(4)=4$ - in which case the function is not continuous at $x=4$, but has a removable singularity at this point. But you only know about the discontinuity if you explore and identify the limit.
First, your use of terminology can be improved. A limit (if it is defined) is just a number, and there is no such thing as a number with a discontinuity. It is your function that may have a discontinuity (and also which may have a limit at$~a$). In the example, the function is given by $x\mapsto\frac{x^2-16}{x-4}$, although it is not explicitly named.
That function is not defined at$~4$, since setting $x=4$ the fraction becomes $\frac00$ which is undefined. It is defined everywhere else, so its domain is $\Bbb R\setminus\{4\}$. To say that a function is continuous or discontinuous at a given point requires that point to be in the domain of the function; therefore you cannot say that you function is discontinuous at$~4$. In fact the function in question is nowhere discontinuous: it is continuous in all points where it is defined, but this excludes the point$~4$.
However for taking the limit of a function$~f$ at a point$~a$, it is not required that $a$ be in the domain of$~f$, just that some neighbourhood of $a$ the result of removing $a$ from that neighbourhood is contained in the domain of$~f$. That's a whole mouth full, but one of the main purposes of limits is to sometimes extract information from a function at points where it is undefined, so the definition must avoid requiring $f(a)$ to be defined. In fact the definition of limit is such that whether or not $f(a)$ is defined is completely ignored: the limit of $f$ at$~a$ depends only on values $f(x)$ for $x$ close but unequal to$~a$.
However, it is a theorem that should $\lim_{x\to a}f(x)$ be defined and the value $f(a)$ as well, then the two are equal if and only if $f$ is continuous at$~a$.
So here's how the reasoning goes. Whenever $x\neq4$ one has $\frac{x^2-16}{x-4}=x+4$, so $x\mapsto x+4$ is a different function from the original one, but one that takes the same values as the original in all points where the original was defined (and in addition the new function is defined at $4$ where the original wasn't). One says that the new function extends the original function. Now since the limit at$~a$ ignores whether a value at$~a$ is defined, one knows for certain that $\lim_{x\to4}\frac{x^2-16}{x-4}=\lim_{x\to4}x+4$ (one side is defined whenever the other is, in which case the two are equal). But $x\mapsto x+4$ is an everywhere continuous function, so in particular $\lim_{x\to4}x+4$ is well defined. Moreover, by the cited theorem this limit must be equal to the value of the new function at$~4$, which value is $4+4=8$. But then also $\lim_{x\to4}\frac{x^2-16}{x-4}=8$ (even though the original function is not defined at$~4$).
You may note that this computation of the limit never even uses the definition of limit, just the knowledge of certain properties of limits and continuity. In fact many limits can be so computed without using (or even knowing) the exact definition of limits. It seems like a swindle, but that is how mathematics often works. (To be sure, one does need to know and use the definition of limits in order to prove the properties that were used.)
Remember the definition of a functional limit: $\lim_{x\to a}f(x) = L$ if and only if for every open neighborhood $V$ of $L$ there exists an open neighborhood $V'$ of $a$ such that $f(V'\setminus \{a\})\subseteq V$. An immediate consequence is that the limit of $f$ at $a$ does not depend on its value at $a$, or in other words, if there is an open neighborhood $V$ of $a$ such that $f(x) = g(x)$ for every point in $V$ except $a$, $f$ and $g$ have the same limit at $a$. In practical terms, this means that when you are evaluating a limit $\lim_{x\to a}f(x)$, you are absolutely allowed to perform algebraic manipulations on $f(x)$ under the assumption that $x\not=a$, but evaluation as $f(a)$ is only valid if you have a theorem saying that it is (that is, saying $f$ is continuous at $a$).
