Basic Limit of $x^2$ Question In Intro To Real Analysis by Bartle & Sherbert there is a discussion / argument that the $\lim _{x \rightarrow c} x^2 = c^2$. As a whole I understand the argument, but I am getting stuck up on two seemingly small details.
For the limit to exist we want to show that  $|x^2-c^2| < \epsilon$ by determining a $\delta$ such that $|x-c|<\delta \rightarrow |x^2-c^2| < \epsilon$.
To do so we note that $ |x^2-c^2| = |x-c||x+c|$. (1) Suppose $|x-c|<1$. Then with a little algebra we find that $|x+c|<2|c|+1$ which implies that $ |x^2-c^2| < |x-c|(2|c|+1)$
so if we let (2) $\delta = \inf [1, \frac{\epsilon}{(2|c|+1)}]$ we end up with our result that $|x-c|<\delta \rightarrow |x^2-c^2| < \epsilon$.
So my questions are : 
1) Why are we allowed to suppose that $|x-c|<1$? I understand that in evaluating a limit we are trying to have $x$ 'approach' a point, but what guarantees that we start so close to our $c$?
2)Whats the purpose of the $1$ in the delta assignment? Im guessing it has to do with our assumption that $|x-c|<1$, but I cant see where it can come into play. It seems that if the latter choice of $\delta$ satisfies the implication even if a large $\epsilon$ is given.
Thanks for the help!
 A: To simplify the above answers:
1) Since we are getting closer and closer to c when we take the limit, surely we will be within 1 of c eventually. Thus we can start our discovery of what $\delta$ must be anywhere we want, and 1 is a good place for our algebra to work out.
2) To do the final step of the proof, that is, make $|x-c||x+c| < \epsilon$, we want to use the fact that $|x-c| < 1$ (in order to ensure that $|x+c|<2|c|+1$) AND that $|x-c|< \frac{\epsilon}{2|c|+1}$, thus we need $\delta$ to precede both values.
A: 1) The general statement goes 'for all $\epsilon>0$, there exists a $\delta>0$ such that...'. You can choose the $\delta$ anyway you want as long as it is $>0$ and satisfies the required $\epsilon$ bound. In particular, you can put restrictions like $\delta < 42$ or any other magic number if it suits your argument. 2) There is nothing special about the $1$ other than it is $>0$. You could have picked any other strictly positive number as long as the subsequent computations are adjusted accordingly.
A: For $1)$, we are allowed to start arbitrarily close to $c$, as long as we close that difference.  For example, we could say $|x-c|<13$, as long as it is positive.  For $2)$, we could pick any positive number, but $1$ just makes the algebra easier.
A: For example, (it doesn't matter much) let $c=0$ and let $\epsilon=10$ (yes, ridiculous). 
It is not true that if $|x-c|\lt \frac{\epsilon}{2|c|+1}$, then $|x^2-c^2|\lt \epsilon$.
For $\frac{\epsilon}{2|c|+1}=10$. Choose $x=4$. Then $|x^2-c^2|=16\gt \epsilon$.
Remark: The estimate (inequality) in the proof you quoted is obtained under the assumption that $x$ is not too far from $c$. So the fact that $x$ is not too far from $c$ has to be built into the choice of $\delta$. It does not matter, all we need to do is to show that for every $\epsilon\gt 0$, there exists a $\delta$ such that $\dots$.  
