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I just realized I did non really internalized $\varepsilon - \delta$ proofs. Here there is an attempt, with some general questions I have.

Proposition: $f(x) := x^2 - 2$ is continuous.


Skratch Work: We assume that $|x - c| < \delta$. Thus, we have

\begin{align} | x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\ & = | x-c| \cdot |x+c|. \end{align}

Assume that $\delta = \frac{c}{2}$. Then we have

$$ |x - c | < \frac{c}{2} \Longleftrightarrow \frac{c}{2} < x < \frac{3c}{2} .$$

By adding $c$ we obtain

$$\frac{3c}{2} < x +c < \frac{5c}{2} .$$

Hence, $x + c < \frac{5c}{2}$, which implies that

$$ | x - c | \cdot | x+ c| < \frac{5c}{2} | x - c| < \varepsilon,$$

if $\delta = \frac{2\varepsilon}{5c}$.


Proof: Let $c$ be arbitrary, and assume that $| x - c | < \delta$. Let $\delta := \min \{ \frac{c}{2} , \frac{2\varepsilon}{5c} \}$. Hence, \begin{align} | x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\ & = | x-c| \cdot |x+c| < \frac{5c}{2} | x - c| < \frac{5c}{2} \delta = \varepsilon. \end{align}

Being $c$ arbitrary, this establishes the continuity of $f(x)$. $\square$


Questions:

  1. Is this proof correct?

  2. I chose the initial value of $\delta$ arbitrarily, but in such a way that depends on the initial choice of $c$ (in doing so I did not follow a proof of this proposition I found, that starts by taking $\delta = 1$). I did this by following a well-known proof of the continuity of $f(x) = \frac{1}{x}$.
    Is this sound?

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    $\begingroup$ You should state that you're assuming that the limit as $x$ approaches $c$ is $c^2-2$. Now you also need to copy your argument from above the argue that $|x+c|<\frac{5c}{2}$, right now that statement is unjustified in the proof. $\endgroup$ – Michael Burr Jun 6 '16 at 0:50
  • $\begingroup$ @MichaelBurr: Thanks a lot for your feedback! Just two question in what follow, because I never saw the steps you mention in these kind of proofs (or I didn't notice them). $\endgroup$ – Kolmin Jun 6 '16 at 1:25
  • $\begingroup$ (1) Where should I write down $\lim_{x \to c} f(x) = c^2 - 2$? I think you should say to put it at the beginning, but isn't it a standard assumption that usually does not show up in these kind of proofs? $\endgroup$ – Kolmin Jun 6 '16 at 1:25
  • $\begingroup$ (2) How can I actually copy the argument concerning $| x + c | < \frac{5c}{2}$ still maintaining the forward structure of the proof? Thanks a lot! $\endgroup$ – Kolmin Jun 6 '16 at 1:27
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Your proof didn't cover the case $c < 0$. I would do it as follows:Let $c \in \mathbb{R}, \epsilon > 0$ be arbitrary, choose $\delta = \min \{1,\frac{\epsilon}{1+2|c|}\}$. Thus if $|x-c| < \delta $, then $$|x-c||x+c| \leq |x-c|(|x-c| + 2|c|)< |x-c|(1+2|c|)< \dfrac{\epsilon}{1+2|c|}\cdot (1+2|c|) = \epsilon .$$

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  • $\begingroup$ Thanks a lot for the answer! Actually I knew this proof, but I could not really make sense of the choice of $1$. Moreover, I don't see why I don't cover the case of $c$. Could you be a bit more explicit? $\endgroup$ – Kolmin Jun 6 '16 at 2:17
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    $\begingroup$ For if $c < 0$, then your $\delta < 0$. $\endgroup$ – DeepSea Jun 6 '16 at 2:18
  • $\begingroup$ Thanks, I see. As mentioned, I still have some questions, and the answer does not really address them. (1) What about the choice of $1$ (the other element of the $\min$ just follows)? (2) Why $1$, without any relation to $c$? (3) Concerning this point, what if I would have chosen $\delta = \frac{ | c |}{2}$? $\endgroup$ – Kolmin Jun 6 '16 at 2:24
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    $\begingroup$ For your first question, you can take any small enough but positive real will do. Question #2: $1$ is a popular choice. Question #3: your $\delta = \dfrac{|c|}{2}$ does not work here since it might be $0$ if $c = 0$, and it is independent of $\epsilon$ which is false. To provide you with a counter example you can first assume $c > 0$, and choose $\theta > 0, x = c+\theta \implies |x^2-c^2| = |(c+\theta)^2 - c^2| = 2c\theta + \theta^2 > 2c\theta > \epsilon$ whenever $\theta > \dfrac{\epsilon}{2c}$. Thus in this case, you don't have continuity at $x = c$. $\endgroup$ – DeepSea Jun 6 '16 at 2:43
  • $\begingroup$ Thanks a lot, that's extremely useful! I would even go as far as to suggest to edit your answer with these comments, because – as it stands – the comments answer the question more than the answer itself. Just one last question: can we say that $1$ is a standard choice when the domain is $\mathbb{R}$, while we start from something like $\frac{c}{2}$ when the domain is restricted (e.g. $x \in (0, \infty)$ like in the case of $f(x) = \frac{1}{x}$)? $\endgroup$ – Kolmin Jun 6 '16 at 2:49

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