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I want to prove the following properties of limits using the $(\delta,\epsilon)$-definition of a limit. I will list each property followed by a proof.

Definitions: Let $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ be two real functions, and assume that for a limit point $c\in\mathbb{R}$, $$\lim_{x\to c}f(x)=L \qquad \lim_{x\to c}g(x)=M$$

  • $$\lim_{x\to c}f^2(x)=L^2$$
    $\color{red}{Proof:}$ From the definitions, we know that for all $\epsilon'>0$,there exists a $\delta>0$ such that $$0<|x-c|<\delta\implies|f(x)-L|<\epsilon'$$ Clearly, the right hand side of the implication yields the implication $$f(x)\in(-\epsilon'+L,\epsilon'+L)$$ Define $\max(|-\epsilon'+L|,|\epsilon'+L|)=\alpha$ and $\min(|-\epsilon'+L|,|\epsilon'+L|)=\beta$. So, without a proof, I state that: $$f^2(x)\in(\beta^2,\alpha^2)*$$ And, define $\epsilon=\alpha^2-L^2$ Clearly, $\epsilon>0$. Now, statement with $*$ implies that $$f^2(x)\in(-\epsilon+L^2,\epsilon+L^2)\\ \implies|f^2(x)-L^2|<\epsilon,\quad\forall\epsilon>0$$

    • $$\lim_{x\to c}f(x)g(x)=L\times M$$ $\color{red}{Proof:}$ From above property, we know that $$\lim_{x\to c}\{f(x)+g(x)\}^2=(L+M)^2\\ \implies\lim_{x\to c}f^2(x)+g^2(x)+2f(x)g(x)=(L+M)^2\\ \implies\lim_{x\to c}f(x)g(x)=L\times M$$

Are these proofs valid?

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  • $\begingroup$ How did you arrive at the universal quantifier $\forall \epsilon > 0$ at the end of the first proof? You're defining $\epsilon$ using $\alpha$! $\endgroup$ – The Vee Nov 15 '16 at 18:15
  • $\begingroup$ And alpha is defined in terms of... $\endgroup$ – codetalker Nov 15 '16 at 18:20
  • $\begingroup$ And as for the universal quantifier $\forall\epsilon>0$, I didn't necessarily proved it, just mentioned it because it is defined in terms of positive real numbers, and as $\epsilon'$ can take any value greater than $0$, it is obvious that we can define $\epsilon$ in terms of $\epsilon'$ $\endgroup$ – codetalker Nov 15 '16 at 18:34
  • $\begingroup$ And that's where the proof would fail a review. Not because it'd be wrong, but incomplete at that point. $\endgroup$ – The Vee Nov 15 '16 at 18:42
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Your proof sounds alright [It's not, see Update below!] but contains way too many unfinished avenues. You need to be explicit about the algorithm "given ε, here's δ" that can actually be performed (using the input you have as assumptions), to convince the reader that the proof works. Here's a bit more elegant proof without relying on "So, without a proof, I state that:" and on supposing that given an $\epsilon$ you could somehow backtrack an $\epsilon'$ from the $\alpha$.

While it's perfectly fine to prove the limit of square first and then extend the validity to the limit of product of functions using $(a+b)^2$, it's usually done the other way. It's less work because $\lim f(x)^2$ is then just a special case of $\lim f(x) g(x)$ for $g = f$.

Knowing that

$$(\forall \epsilon>0)(\exists \delta>0)(\forall x, |x-c| < \delta)(|f(x) - l| < \epsilon)$$

and

$$(\forall \epsilon>0)(\exists \delta>0)(\forall x, |x-c| < \delta)(|g(x) - m| < \epsilon),$$

you can expand

$$f(x)g(x) - lm = (f(x) - l)g(x) + l(g(x) - m)$$

and limit that from above as:

$$|f(x)g(x) - lm| ≤ |f(x) - l| |g(x)| + |l| |g(x) - m|.$$

Let's assume a particular $\epsilon > 0$ that's smaller than both $|l|$ and $|m|$. Note that this could only be done if $lm \ne 0$, but the case of one of the limits being zero is easy and I'm not going to do that. We can then estimate

$$|f(x)g(x) - lm| ≤ \epsilon(|g(x)| + |l|).$$

Now

$$|g(x)| = |(g(x) - m) + m| ≤ |g(x)-m| + |m| ≤ \epsilon + |m| ≤ 2|m|$$

and finally

$$|f(x)g(x) - lm| ≤ \epsilon(2|m| + |l|).$$

So, if someone gives me an $\epsilon' > 0$, ($\color{red}{\forall \epsilon' > 0}$) I can find define my own

$$\epsilon := \min\left\{ \frac{\epsilon'}{2|m| + |l|}, |l|, |m| \right\}$$

find a suitable $\delta$ for that because I know one exists (using the verbatim version of the assumptions listed in the beginning), ($\color{red}{\exists \delta > 0}$) and using the two claims I started with prove that $\color{red}{\forall x, |x-c| < \delta}$,

$$\color{red}{|f(x)g(x) - lm| ≤} \epsilon(2|m| + |l|) = \frac{\epsilon'}{2|m| + |l|}(2|m| + |l|) = \color{red}{\epsilon'},$$

QED.

Update: Your proof actually suffers of the same special case of $L = 0$ and that illustrates how a fault can easily hide beyond any incompleteness. In this case, you know that $f(x) \in (-\epsilon', \epsilon')$ but you go on to define $\alpha = \max\{|-\epsilon'|,|\epsilon'|\} = |\epsilon'|$ and likewise $\beta$. But certainly we're not given that $f^2(x) \in (|\epsilon'|^2, |\epsilon'|^2)$... in fact, that interval is invalid! (But it's not necessarily an element even if you made it closed.)

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  • $\begingroup$ I completely agree with you, @The Vee. The whole point of me doing it the reverse way was because it could be done that way, and partially because I thought it was a little easy. +1 $\endgroup$ – codetalker Nov 15 '16 at 18:42
  • $\begingroup$ Thank you! By the way, I think I just succeeded in finding an actual flaw in your proof, check back in a few minutes. $\endgroup$ – The Vee Nov 15 '16 at 18:43
  • $\begingroup$ Thnx a lot @The Vee, I will handle the special case somehow:) $\endgroup$ – codetalker Nov 15 '16 at 19:07
  • $\begingroup$ except that, is there any fault that you see? $\endgroup$ – codetalker Nov 15 '16 at 19:07
  • $\begingroup$ See my last edit. I had to bound how my $\epsilon$ was retrieved from the $\epsilon'$ (but it can still be done for any $\epsilon'>0$). I think the same will be needed in yours; actually, it's a generalization of the phenomenon that causes the above error. $\endgroup$ – The Vee Nov 15 '16 at 19:16

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