# prove $x \mapsto x^2$ is continuous

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.

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 $$\left|x+x_0\right|= \left|x-x_0+2x_0\right|\le \left|x-x_0\right|+\left|2x_0\right|<\delta+2\left|x_0\right|$$ Thus, $$\left|x-x_0\right|<\delta\implies \left|f(x)-f(x_0)\right|=\left|x+x_0\right|\left|x-x_0\right|<(\delta+2\left|x_0\right|)\delta$$ 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

• it is getting bigger than $\epsilon$ if i take $\frac{\epsilon}{2}$ Dec 23 '12 at 12:28
• @doniyor I updated my answer with the last details Dec 23 '12 at 13:06
• thanks a lot. it was a little bit confusing the whole topic Dec 23 '12 at 13:25
• Can I ask why did you add one half at the end of the proof ?
– VLC
Apr 12 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,$$|f(x)-f(x_0)|=|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|x-x_0||x+x_0|\leq |x-x_0|(1+2|x_0|)<(1+2|x_0|)\delta<\epsilon.$$