I have one elementary doubt on the proof that $f(x)=x^2$ is continuous for every $a \in \Bbb R$. The initial steps usually presented are: to deduce which $\delta$ will work we write:


Now, we must find some bound for $|x+a|$, so we require that $|x-a|<1$ or some other positive real number. Here comes my first doubt, this trick appears very often, what's the intuition on making $|x-a|<1$?

After that we have that this implies $x < 1 + a$ and this implies that $|x+a| < 1 + 2|a|$. Now it's obvious that if we set $|x-a|< \varepsilon/(1+2|a|)$ and this obviously will make $|f(x)-f(a)|<\varepsilon$. However this was deduced under the hypothesis that $|x-a|<1$, so what if it were not true that $|x-a|<1$? Also the end of the proof is normally saying that we must take

$$\delta = \min\{1, \varepsilon/(1+2|a|)\}$$

However it happens that if the minimum is $1$, it seems to not work, because when we use this $\delta$ the $\varepsilon$ doesn't even appear. I think my problem is that I didn't really get the trick of making $|x-a|<1$.

Thanks very much in advance!


The number $1$ is arbitrary; it was chosen for its simplicity. You are right that $|x-a| < 1$ is not satisfied for every $x$, but it does not matter. We are only interested in what happens when $x$ is very close to $a$; more precisely, when $x$ satisfies $|x-a| < \delta$. But by our choice of $\delta$, $|x-a| < 1$ is always satisfied whenever $x$ is $\delta$-close to $a$.

  • $\begingroup$ Hi @AlexP, thanks very much! Now I've got it, if $|x-a|$ is less than the minimum, it is less than both, and this will assure that for every $\varepsilon$ the condition holds. Thanks very much for your help! $\endgroup$ – user1620696 May 27 '13 at 0:36

The thing is that we really care about $|x-a|$ being very small, even close zero, so we may as well assume that it is smaller than $1$. Because given an $\varepsilon$ we need to tell what sort of $\delta$ suits it, we can choose it to our liking.

And since we are free to choose our $\delta$, we can always choose $\delta$ to be smaller than $1$, e.g. $\min\left\{\frac12,\frac\varepsilon{1+2|a|}\right\}$.

  • 1
    $\begingroup$ using "min" is very helpful $\endgroup$ – Stefan Smith May 27 '13 at 1:01

If some particular $\delta$ works for the given $\epsilon$, i.e., if for all $x$ satisfying $0 < |x - a| < \delta $ we have $|f(x) - L| < \epsilon$; then any smaller $\delta' < \delta$ also works for $\epsilon$. This is because $\{x \, :\, 0< |x - a| < \delta' \}$ is a subset of $\{x \, : \, 0< |x - a| < \delta \}$ when $\delta' < \delta$.

So we may take $\delta$ as small as possible as we want, everytime we need to put an upperbound on an expression involving $x$. Then, we can set $\delta$ to be the minimum of the upperbounds for $\delta$.


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