My question regards using the $\delta$-$\epsilon$ definition of a limit to prove that:

$\lim_{x\to2} x^2=4$

Now, in the book I'm working out of, I can follow almost all of the arguments made in the proof. The proof unfolds as follows:

Assume $\epsilon > 0$, we will show that there exists a $\delta > 0$ such that $\vert x^2-4 \vert < \epsilon$ whenever $0 < \vert x - 2 \vert < \delta$.

We see that $\vert x^2 - 4 \vert = \vert x -2 \vert \cdot \vert x + 2 \vert$. Then for all $x\in (1,3)$ it follows that $x+2 < 5$ and thus $\vert x + 2 \vert < 5$. So letting $\delta$ be the minimum of $\epsilon/5$ and $1$, it follows that, if $0 < \vert x - 2 \vert < \delta$, then

$$\vert x^2 - 4 \vert = \vert x - 2 \vert \cdot \vert x + 2 \vert < \frac{\epsilon}{5}\cdot 5 = \epsilon.$$

But why do we say that $\delta$ is the minimum of $\epsilon/5$ and $1$? Why don't we just let $\delta = \epsilon/5$? I'm not quite sure why we mention $1$ here.

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    $\begingroup$ Because we need to have a bound for $|x+2|$, of which it depends on the assumption that $|x-2|<1$. $\endgroup$ – Juniven Feb 23 '17 at 16:25

Because $\varepsilon$ could be any positive number.

If for example, $\varepsilon=20$, then $\delta=\varepsilon/5$ does NOT work. Say $x=5$, then $|x-2|=3< 4=\varepsilon/5$, but $|x^2-4|=21>\varepsilon$!

Note. If we have a function $f:\mathbb R\to\mathbb R$, which we need to show that it is continuous at $x_0$, it is usually convenient to pick an initial interval $(x_0-d,x_0+d)$, where we look for $\delta=\delta(\varepsilon)$. And as $x\in(x_0-d,x_0+d)$, then clearly, $\delta(\varepsilon)\le d$.

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  • $\begingroup$ I'm not familiar with the notation $\delta = \delta(\epsilon)$, can you clarify what your note describes? $\endgroup$ – ClownInTheMoon Feb 24 '17 at 21:52
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    $\begingroup$ I simply want to point out that $\delta$ depends in general on $\varepsilon$. $\endgroup$ – Yiorgos S. Smyrlis Feb 24 '17 at 22:45

If you take the max, then there is no guarantee that you will be within the $\delta$ range you desire. Since we have $|x-2|$ and $|x+2|$, taking the min of the epsilon values will guarantee that we are within $\delta$ of $2$.

Go here:


Copy my settings and play around with it to make sure you understand, it's very helpful to see it visually.

enter image description here

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