Just to elaborate on Michael Hardy's answer a little bit...
Suppose we have a rational function written like $f(x)=\frac{p(x)}{q(x)}$. Performing the long division $p(x)\div q(x)$ gives you a different way of writing the same function, $$f(x)=a(x)+\frac{r(x)}{q(x)},$$ as a polynomial plus a rational function whose numerator's degree is smaller than its denominator's degree. (To be more specific, $a(x)$ has degree equal to the degree of the original numerator $p(x)$ minus the degree of the denominator $q(x)$. The degree of $r(x)$ is smaller than that of $q(x)$ because $r(x)$ is the remainder of the long division.)
As you already understand, $\frac{r(x)}{q(x)}$ tends to zero as $x$ tends to infinity, because the degree of the numerator is smaller than the degree of its denominator. That means that as $x\rightarrow\infty$, the function $f(x)$ tends towards the polynomial $a(x)$.
If $p(x)$ has degree exactly one more than that of $q(x)$, then $a(x)$ has degree one: it's a line. We call this line a slant or oblique asymptote of $f(x)$. If the degree of $p(x)$ is more than one higher than that of $q(x)$, then $a(x)$ is a quadratic, or a cubic, or whatever, but it's not a line. So $f(x)$ doesn't have a slant asymptote in this direction. (We still might say something like $f(x)$ asymptotically approaches a(x).)
Hopefully this explains why asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator. It also might give you a hint for how you can find slant asymptotes of functions that aren't rational: if you can rewrite your function as a line plus something that goes to zero, you've got yourself an asymptote!
This is an answer on the question from the khanacademy.org
Full credit for this answer goes to the user "Tronax" from khanacademy.org:
I will try to show the difference between the two types of what we previously called "undefined".
Let's start with the "removable discontinuity", which actually says, that the graph at the point of x could be absolutely any y! And there is no way to find it. It could be 0, -7, 65, 23/7, 3982.3 etc. Math guys agreed between themselves, that if they can't understand it better and find its value - it does not exist. So now we just draw an empty circle on the curve of our graphs in spots that are known to have a removable discontinuity.
This happens if the function is of a form that includes the same non constant factor in both numerator and denominator. For example y = (x+1)(x+2) / (x+1). Notice (x+1) factor is both on the numerator and on the denominator.
For x=-1, (x+1) equals zero, so we say that at x=-1 function has a removable discontinuity.
But... why?
Because when you input x=-1 and try to solve for y:
y = (-1+1)(-1+2) / (-1+1)
y = 0*(-1)/0
y = 0/0
or (this is one of interpretations):
0y = 0
You could substitute any number into y, this expression will fit them all, since anything multiplied by 0 equals 0.
This is removable discontinuity. The graph around the point of it, looks just like it would, if there was no removable discontinuity.
The second type is the "vertical asymptote". It occurs when for some x, the denominator (and only denominator) equals zero. It's somewhat easier to understand. Let's think about what happens when we see 8/4. This says: how many times, can we fit the denominator 4 into 8? It takes 2 times. For 8/2 it takes 4 times; for 8/1 - 8 times; 8/0.5 - 16 times; 8/0.25 - 32 times. The lower and closer to zero our denominator gets, the more times we can fit it into the same 8. When it becomes really really small, it can take millions of times! When it gets such a small number that we can barely distinguish it from zero, it seems like it would take almost infinite times of that number to fit our 8! That's why when denominator approaches 0, we say that the result approaches infinity.
When x approaches 0, y=8/x approaches infinity and y=-8/x approaches negative infinity.
Don't be confused, when x actually equals zero, y=8/x is undefined! In this case, you could say that even infinity is not enough :)
When saying "approaches zero", we only mean that it gets very very very close to zero, but not zero.
"Vertical asymptote at x = k" only refers to what happens before and after k on the graph. x = k itself is undefined.
The graph before and after vertical asymptote quickly gets almost parallel to the y-axis.
Hope this is helpful.
Source: https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/discontinuities-of-rational-functions/v/discontinuities-of-rational-functions
Best Answer
Let's be a bit careful with the term undefined, since it has a bit more of a precise meaning than the one you've used in your question.
You correctly observe that $f(-4) = \frac 0 0$ and $f(-6) = \frac{g(x)}0$. Why are they different? The reality is that $\frac 0 0 $ is actually indeterminate, meaning that it does not have an established value, instead of being undefined, which would imply having no value.
Hence, the case of $\frac 0 0 $ is interesting because it implies that there could exist a whole range of values that a function that approaches such a fraction could approach, since it all depends on how fast the numerator and denominator approach 0. In other words, removing the terms that take the function to zero in essence is us recognizing that these terms approach zero at the same rate, and at all points near -4, we can ignore them.
On the other hand, approaching a fraction of $\frac {g(x_0)}0$ where $g(x_0)$ is non-zero means that the function must approach either $\pm\infty$, since there are no ways for any value to exist and be equal to that fraction.
This is some rough intuition, but should shine some light on the difference between removable discontinuities and asymptotes.