Question Regarding Limits and Confusing Notation My calculus teacher assigns us online homework to do.  He never went over any question that looks like this (he in fact said we shouldn't be concerned with this):
Yet, I need to answer this right to progress with my homework.  It stinks because if I get it wrong, I lose points on my homework average.
So, could someone help explain to me what's going on here, and perhaps guide me to a point where I can try to figure out the solution myself?  I'm not asking for a straight answer (although if thats what you want to provide, go for it [since I was told I don't have to know this stuff]), but this stuff really confuses me.  Thank you for your help.
Oh, and in case you need it, here's the original prompt (the question I posted above is just a part of a series of questions that go along with this prompt):

 A: I think the question is quite confusingly worded. It took me several minutes to figure out what it meant -- and it's not as if I don't know the subject matter well.
What must be going on is that you're supposed to imagine reading something like this in a proof:

bla bla bla, and therefore we know that $f$ is continuous, and that $f(c)\ne 0$. We can then apply the definition of continuity with $\varepsilon = $ _______ to find a $\delta$ such that $f(x)$ has the same sign as $f(c)$ for every $x\in(c-\delta,c+\delta)$. Thus, bla bla bla

One of $|f(c)|$ and $|c|$ will make this into a valid argument if you fill it into the blank, and one will produce nonsense. Your task is to select the valid one.
In order to answer the question you need (1) to remember the definition of continuity, and (2) to be able to distinguish a nonsense argument from a valid one. The second of these abilities is often considered too advanced a skill to demand of pre-university students these days (they're supposed to be satisfied with accepting the teacher's judgement in each case), which is probably why your teacher is not allowed to say you must be able to do it...
A: Without loss of generality we may assume that $f(c)$ is positive. 
Let $\epsilon =f(c)$.  By the definition of continuity of $f$ at $c$, there is a $\delta>0$ such that if $|x-c|<\delta$ (and $a\lt c-\delta$, and $c+\delta \lt b$, to make sure we stay in our interval) then $|f(x)-f(c)|<\epsilon$.
Now if you have some experience with inequalities, you should be able to reach the conclusion.
If the "without loss of generality" is not persuasive, if $f(c)<0$, let $g(x)=-f(x)$, apply the above argument to $g(x)$, and see what this says about $f(x)$. 
I did not use precisely the language of the question. But you should now be able to see enough of what is going on to be able to answer that question.
Comment: It will be helpful to draw a picture while figuring out what the second paragraph is saying. 
