I am to show the continuity of this function with the help of $\epsilon$-$\delta$ argument.

The function is: $g: \Bbb{R} \rightarrow \Bbb{R}$, $x \mapsto x^2$.

Given the $\epsilon$-$\delta$ definition of limit, I tried as follows:

We must have: $|f(x)-f(x_0)|<\epsilon$, so then $|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|$, so $|x-x_0|<\frac{\epsilon}{|x+x_0|}$. So now I will choose $\delta=\frac{\epsilon}{|x+x_0|}$.

Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively.


2 Answers 2


No. $\delta$ must only depend on $x_0,\epsilon$ and never on $x$. Here is what we will do:

The problematic term is $\left|x+x_0\right|$. We have that \begin{equation} \left|x+x_0\right|= \left|x-x_0+2x_0\right|\le \left|x-x_0\right|+\left|2x_0\right|<\delta+2\left|x_0\right|\end{equation} Thus, \begin{equation}\left|x-x_0\right|<\delta\implies \left|f(x)-f(x_0)\right|=\left|x+x_0\right|\left|x-x_0\right|<(\delta+2\left|x_0\right|)\delta\end{equation} We must choose a $\delta$ so that $$(\delta+2\left|x_0\right|)\delta<\epsilon$$ Choosing a $\delta$ by the above might be complicating. But because $\delta$ is ours to choose we can simplify things a bit by demanding $\delta<1$. Then, $$(\delta+2\left|x_0\right|)\delta<(1+2\left|x_0\right|)\delta$$ and so it suffices to choose $0<\delta<1$ so that $$(1+2\left|x_0\right|)\delta<\epsilon$$ Things should be straighforward now.

For the sake of completion we have $$(1+2\left|x_0\right|)\delta<\epsilon\iff \delta<\frac{\epsilon}{1+2\left|x_0\right|}$$ But because $\delta<1$ we must choose $\delta>0$ so that $$\delta<\min\left\{1,\frac{\epsilon}{1+2\left|x_0\right|}\right\}$$ Taking $$\delta=\frac12\min\left\{1,\frac{\epsilon}{1+2\left|x_0\right|}\right\}$$ completes the proof

  • $\begingroup$ it is getting bigger than $\epsilon$ if i take $\frac{\epsilon}{2}$ $\endgroup$
    – doniyor
    Commented Dec 23, 2012 at 12:28
  • 1
    $\begingroup$ @doniyor I updated my answer with the last details $\endgroup$
    – Nameless
    Commented Dec 23, 2012 at 13:06
  • $\begingroup$ thanks a lot. it was a little bit confusing the whole topic $\endgroup$
    – doniyor
    Commented Dec 23, 2012 at 13:25
  • 4
    $\begingroup$ Can I ask why did you add one half at the end of the proof ? $\endgroup$
    – VLC
    Commented Apr 12, 2021 at 13:41

Assume we found $\delta$,

$|x|\leq |x-x_0+x_0|\leq|x-x_0|+|x_0|<\delta+|x_0|.$

If this $\delta$ works, then any $\delta'\leq \delta$ also works, therefore we may assume $\delta<1$ as well. Now we have an estimate of the form $|x|<1+|x_0|$. One can see that we just restricted $x$ in the open interval $(x_0-1,x_0+1)$.

Now, we have

$$|f(x)-f(x_0)|=|x^2-x_0^2|\leq (1+2|x_0|)|x-x_0|$$.

One can observe that $\delta<\frac{\epsilon}{(1+2|x_0|)}.$

Since we also want to estimate first, namely $|x|<1+|\delta|$. Therefore we choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$

Coming to the formal proof;

Fix $x_0\in\Bbb{R}$, given $\epsilon>0$. Choose $\delta<\text{min}\{1,\frac{\epsilon}{(1+2|x_0|)}\}.$

Let $x\in\Bbb{R}$ be such that $|x-x_0|<\delta$. Then $|x-x_0|<1, |x|=|x-x_0+x_0|\leq |x-x_0|+|x_0|<1+|x_0|$.

Now,$|f(x)-f(x_0)|=|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|x-x_0||x+x_0|\leq |x-x_0|(1+2|x_0|)<(1+2|x_0|)\delta<\epsilon.$


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