# 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!

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You can start as close to c as you wish, since you consider the limit as $x\rightarrow c$. By putting $1$ in the $\delta$ assignment as above, you make sure you only consider $x$ such that $|x-c|<1$. Then you are allowed to suppose $|x-c|<1$ and you can perform the necessary algebra. – 1015 Jan 8 '13 at 1:51
Here is a related problem. – Mhenni Benghorbal Jan 8 '13 at 3:32

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 $\dots$.

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I appreciate the example you provided, but please confirm : surely you meant $\frac{\epsilon}{2|c|+1}=10$, correct? The rest of the result still holds it seems. – MSEoris Jan 8 '13 at 2:54
@MSEoris: Thanks for careful reading. Error of arithmetic corrected. Happily it makes no difference. One could produce a similar example for any $c$. – André Nicolas Jan 8 '13 at 3:15

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.

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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.

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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.

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