2
$\begingroup$

I am trying to understand the epsilon delta definition of continuity but every example I look at they seem do do something I don't agree with. Example. Show that $f(x) = x^{2} - 2$ is continuous at some point $x = x_{0}$.

Now $\forall \epsilon>0, \exists \delta>0, \text{ s.t. } |x - x_{0}| < \delta \implies |(x^{2}-2)-(x_{0}^{2}-2)|<\epsilon$

Now notice $$\begin{align}|(x^{2}-2)-(x_{0}^{2}-2)| &= |x^{2}-x^{2}_{0}|\\ &= |x^{2} - 2xx_{0} + x_{0}^{2} - 2x_{0}^{2} + 2xx_{0}| \\ &= |(x-x_{0})^{2} + 2x_{0}(x - x_{0})|<\epsilon\end{align}$$

I understand and agree with all of the above, now the step that seems a bit off to me:

$$\le |(x - x_{0})|^{2} + |2x_{0}(x-x_{0})| < \epsilon$$ by the triangle inequality. But we don't know this is less than epsilon since we now have something that is greater than what we actually know is less than epsilon, this comes up time and time again in many proofs and I can't get my head around why we say this is still less than epsilon.

$\endgroup$
3
$\begingroup$

I'm assuming your true concern is understanding $\epsilon$,$\delta$-proofs. We first go over the definition of continuity at a point, then give an example.

The Definition

Let $f$ be a function from some subset of $\mathbb{R}$, call it $D$ for domain, to another subset of $\mathbb{R}$, call it $C$ for codomain. That is,$$f:D\subseteq\mathbb{R}\to C\subseteq\mathbb{R}$$ Let $a$ be an arbitrary element of the domain. That is, $$a\in D$$ $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in D$, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.

An Example

Let $f:[3,\infty)\to \mathbb{R}$ be defined by $$f(x)=\sqrt{2x-6}$$ for every $x\in[3,\infty)$. (It is important to define the function explicitly; that is, the domain is relevant for this proof.)

$f$ is continuous at $4$.

Proof. To prove $f$ is continuous at $4$, we will show that for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$.

Let $\epsilon>0$ be arbitrary. Notice for each $x\in[3,\infty)$, $$|f(x)-f(4)|=\sqrt{2}|\sqrt{x-3}-1|=\sqrt{2}\left|\frac{x-4}{\left(\sqrt{x-3}\right)+1}\right|<\sqrt{2}|x-4|\tag{1}$$ Define $\delta\overset{\text{def}}{=}\dfrac{\epsilon}{\sqrt{2}}$. It is important to understand why we defined $\delta$ this way (in relation to $\epsilon$). Anyway, we continue with the proof. For each $x\in[3,\infty)$, if $|x-4|<\delta$, then $$|f(x)-f(4)|\overset{(1)}{<}\sqrt{2}|x-4|<\sqrt{2}\cdot\delta=\epsilon$$ Recall $\epsilon>0$ was arbitrary. Therefore for each $\epsilon>0$, there exists $\delta>0$ (namely $\epsilon/\sqrt{2}$), such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$. Therefore $f$ is continuous at $4$.$\square$

$\endgroup$
  • $\begingroup$ You have done the same thing again.. $|f(x) - f(4)|<\epsilon$ but you have now said $|f(x) - f(4)|<\sqrt{2}|x-4|$. So we know $|f(x) - f(4)|<\epsilon$ but now we have that $|f(x) - f(4)|<\sqrt{2}|x-4|$ so you don't know if $\sqrt{2}|x-4| < \epsilon$ since it is greater than $|f(x)-f(4)|$. How does nobody else see this? $\endgroup$ – user2850514 Apr 30 '14 at 21:40
  • $\begingroup$ Understand "$|f(x)-f(4)|<\epsilon$" is a consequence of the following three facts: $|f(x)-f(4)|<\sqrt{2}|x-4|$ and $|x-4|<\delta$ and $\delta=\epsilon/\sqrt{2}$ $\endgroup$ – Alberto Takase Apr 30 '14 at 21:49
  • $\begingroup$ Do you not believe me when I say for each $\epsilon>0$, there exists a $\delta>0$ such that IF $|x-4|<\delta$, THEN $|f(x)-f(4)|<\epsilon$? $\endgroup$ – Alberto Takase Apr 30 '14 at 22:00
  • $\begingroup$ Right so initially we don't actually have $|f(x) - f(4)|<\epsilon$ in fact we just have $|f(x)-f(4)|$ and we need to use $\delta$ to get $\epsilon$? So if $|f(x)-f(4)| < \sqrt{2}\delta$.. now we let $\epsilon = \sqrt{2}\delta$ ie. $\delta = \epsilon/\sqrt{2}$.. I think I understand now. $\endgroup$ – user2850514 Apr 30 '14 at 22:03
  • $\begingroup$ Your comment "we need to use $\delta$ to get $\epsilon$" is absolutely correct! I am glad you are beginning to see the importance of $\delta$. The main idea is to be able to choose a "delta" from any "epsilon." How is this done? By defining "delta" in terms of the given "epsilon." $\endgroup$ – Alberto Takase Apr 30 '14 at 22:06
3
$\begingroup$

If what you presented is indeed the entire example, then something is definitely off, because no $\delta$ has been given that makes the inequality true. Take a look again and see if there's a sentence that says something along the lines of "Let $\delta =\ldots$," followed by a possibly funny formula in $\epsilon$ and $x_0$.

Added in reply to OP's comments: Without seeing exactly how the book (or whatever you're studying from) presents things, it's a little hard to be sure where the difficulty is, but I'll take a stab at it.

How I think the inequalities should be presented is something like:

$$\begin{align}|(x^{2}-2)-(x_{0}^{2}-2)| &= |x^{2}-x^{2}_{0}|\\ &= |x^{2} - 2xx_{0} + x_{0}^{2} - 2x_{0}^{2} + 2xx_{0}| \\ &= |(x-x_{0})^{2} + 2x_{0}(x - x_{0})|\\ &\le |(x - x_{0})|^{2} + |2x_{0}(x-x_{0})|\\ &\lt\delta^2+|2x_0|\delta\\ &\le\delta(1+|2x_0|)\\ &\le\epsilon\end{align}$$

But all this should be preceded by the specification

$$\delta= \begin{cases} {\epsilon\over1+|2x_0|}& \epsilon\lt1+|2x_0|\\ 1 &\epsilon\ge1+|2x_0|\\ \end{cases}$$

which can also be written as

$$\delta=\min\{1,{\epsilon\over1+|2x_0|}\}$$

The justifications for the inequalities ending in "$\lt\epsilon$" is the triangle inequality for the first one (i.e., $|a+b|\le|a|+|b|$ for any $a$ and $b$), the assumption $|x-x_0|\lt\delta$ for the second, the specification $\delta\le1$ for the third, and the specification $\delta\le\epsilon/(1+|2x_0|)$ for the finale. (You might note there's only one strict inequality in there. You could make the final two strict as well by asking that $\delta$ be strictly less than the minimum of $1$ and $\epsilon/(1+|2x_0|)$, but you don't need to.)

A difficulty beginners often have (I know I had it, at least), is understanding that a key step is coming up with a specification for $\delta$ that makes the inequalities ending in the "$\lt\epsilon$" work. What I sometimes do (or did) is to go ahead and writing down the inequalities that I want to be true, and then ask myself what's the simplest thing I can say about $\delta$ that makes them true. Once I've figured that out, I go back and rewrite things starting with the $\delta$.

I hope this helps.

$\endgroup$
  • $\begingroup$ Of course there is more but I want to understand this step first. $\endgroup$ – user2850514 Apr 30 '14 at 20:40
  • 1
    $\begingroup$ @user2850514, but you've left out one of the most important steps. You need to find where they've specified the $\delta$. $\endgroup$ – Barry Cipra Apr 30 '14 at 20:44
  • $\begingroup$ Well I don't understand this step so I won't understand the rest until this is sorted.. The rest says $< \delta^{2} + 2|x_{0}|\delta = \delta(\delta + 2|x_{0}|)$. So for $\delta < 1$: $\delta^{2} < \delta \implies |x^{2} - x_{0}^{2}| < \delta(1 + 2|x_{0}|)$. We want this to read as $|x^{2} - x_{0}^{2}| < \epsilon$ so we choose $\delta(1 + 2|x_{0}| = \epsilon)$ hence $\delta = \frac{\epsilon}{1 + 2|x_{0}|}$. When $\delta < 1, \epsilon < 1 + 2|x_{0}|$. $\delta = 1$ gives $1+2|x_{0}| \le \epsilon$. This means that for all $\epsilon \ge 1+2|x_{0}|$, $\delta = 1$ will satisfy the definition. $\endgroup$ – user2850514 Apr 30 '14 at 20:57
  • $\begingroup$ So $$\delta = \begin{cases}\frac{\epsilon}{1 + 2|x_{0}|} & \epsilon < 1 + 2|x_{0}| \\ 1 & \epsilon \ge |1 + 2|x_{0}| \end{cases}$$ $\endgroup$ – user2850514 Apr 30 '14 at 20:59
1
$\begingroup$

Because of exactly the problems you are having, you should do as I was taught in Calculus I here: Whip out a piece of scratch paper, and as you do work forward from $\lvert x - x_0 \rvert < \delta$ to get (as you do): $$ \lvert (x - x_0)\rvert^2 + \lvert 2x_0 \rvert \cdot \lvert x - x_0\rvert < \delta^2 + 2 \lvert x_0 \rvert \delta $$ This tells you that for that $\delta$ you need $\epsilon$ at least this, i.e., $$ \delta^2 + 2 \lvert x_0 \rvert \delta \le \epsilon \\ $$ As $\delta$ is supposed to be small, there is no harm in assuming $\delta \le 1$, which makes $\delta^2 + 2 \lvert x_0 \rvert \delta \le \delta (1 + 2 \lvert x_0 \rvert)$ so you pick: $$ \delta = \min\left( 1, \frac{\epsilon}{1 + 2 \lvert x_0 \rvert} \right) $$ OK, now to clean work.

Pick $\epsilon > 0$, and take $\delta = \min(1, \epsilon / (1 + 2 \lvert x_0 \rvert))$. Now, if: $$ \lvert x - x_0 \rvert < \min\left(1, \frac{\epsilon}{1 + 2 \lvert x_0 \rvert}\right) \\ $$ then, using the fact that by our selection of $\delta$ we have $\lvert x - x_0 \rvert \le 1$: \begin{align} \lvert (x^2 - 2) - (x_0^2 - 2) \rvert &= \lvert x^2 - x_0^2 \rvert \\ &= \lvert x^2 - 2 x x_0 + x_0^2 - 2 x_0^2 + 2 x x_0 \rvert \\ &= \lvert (x - x_0)^2 + 2 x_0 (x - x_0) \rvert \\ &\le \lvert x - x_0\rvert^2 + \lvert 2 x_0 (x - x_0) \rvert \\ &\le \lvert x - x_0 \rvert + \lvert 2 x_0 \rvert \cdot \lvert x - x_0 \rvert \\ &= \lvert x - x_0 \rvert \cdot \lvert 1 + 2 \lvert x_0 \rvert \rvert \\ &\le \lvert x - x_0 \rvert \cdot ( 1 + 2 \lvert x_0 \rvert ) \\ &\le \epsilon \end{align} Esteban Herreros (my professor in first year) threatened to fail any of us who made the secret for clean limit proofs known outside of our class. So please don't tell anybody.

$\endgroup$
  • $\begingroup$ That's great to know, thank's a lot. $\endgroup$ – user2850514 Apr 30 '14 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.