Could it be anything?

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Two young mathematicians investigate the arithmetic of large and small numbers.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Hey Riley, I’m having a little trouble with this limit $\lim _{x\rightarrow 0}\frac {9x^2-4x}{3x^2-x}?$
Riley: Well, just plug in $0$ and see what you get, right?
Devyn: I tried that, here’s what I got... $\eval {\frac {9x^2-4x}{3x^2-x}}_{x=0} = \frac {9(0)^2-4(0)}{3(0)^2-(0)}=\frac {0}{0}$
Riley: Hm, so that... equals 1 right? I mean, anything divided by itself must be 1?
Devyn: I was thinking the same thing. Initially. But then I tried simplifying stuff first and look what happened. $\lim _{x\rightarrow 0}\frac {9x^2-4x}{3x^2-x}=\lim _{x\rightarrow 0}\frac {x(9x-4)}{x(3x-1)}=\lim _{x\rightarrow 0}\frac {9x-4}{3x-1}=\frac {9(0)-4}{3(0)-1} = \frac {4}{1} = 4$
Riley: Ok... so the answer is 4? Or is it 1? It can’t be both right?
Devyn: Right, that’s why I’m stuck.
Riley: Well, we originally thought it was 1 because it was “a number divided by itself.” But that number was zero, and dividing by zero is usually where things go wrong in math.
Devyn: Good point. Come to think of it, I can’t just plug in the limit value because that only works when the function is continuous at that point. And I can’t evaluate the top and bottom separately like that, because that limit law only works when the bottom isn’t zero. Crap, I really messed that one up, the limit must be $4$ then. Thanks!
Devyn: ...
Devyn: Also, uh, don’t mention this conversation to anybody ok?
Riley: What conversation?
Devyn: Exactly.

1 : Consider the function $f(x) = \frac {x}{x}.$ $f(0) = \answer {DNE}\qquad \lim _{x\to 0} f(x) = \answer {1}.$

2 : Consider the function $f(x) = \frac {4x}{x}.$ $f(0) = \answer {DNE}\qquad \lim _{x\to 0} f(x) = \answer {4}.$

3 : Consider the function $f(x) = \frac {x}{-3x}.$ $f(0) = \answer {DNE}\qquad \lim _{x\to 0} f(x) = \answer {-1/3}.$