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I'm reading a few proofs of the squeeze theorem to try to understand it and i always find the same inconsistency, for example on this proof on Wikipedia https://en.wikipedia.org/wiki/Squeeze_theorem#Proof

on this particular step:

knowing that

1) $|x-a|<\delta_1 => |g(x)-L|<\epsilon$

2) $|x-a|<\delta_2 => |f(x)-L|<\epsilon$

3) $g(x)-L<h(x)-L<f(x)-L$

we choose a $min(\delta_1,\delta_2 )$ then it holds that

$$-\epsilon < g(x)-L<h(x)-L<f(x)-L$ < \epsilon$$

Ok, so my question is, on lines 1) and 2) why we have the same $\epsilon$? wouldn't it be different for a different function?, even after taking $min(\delta_1,\delta_2 )$, why would both $\epsilon_1$ and $\epsilon_2$ be equal so that you can combine them on a single expression later on?

1) $|x-a|<min(\delta_1,\delta_2 ) => |g(x)-L|<\epsilon_1$

2) $|x-a|<min(\delta_1,\delta_2 ) => |f(x)-L|<\epsilon_2$

let me know if any clarification is needed.

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  • $\begingroup$ Because epsilon(s) can be anything we want. The deltas are detrermined by the epsilon(s). So we what both deltas to work for the one epsilon we chose. $\endgroup$ – fleablood Apr 3 '17 at 18:08
  • $\begingroup$ Consider the two statements. "For any $x$ there is a value, $x^2$" and "for any $x$ there is a value $\sqrt[3]{x}$". So I say "hmmm, if $x=7$ I can find $7^2$". Okay. "And for $x=7$ I can find $\sqrt[3]{7}$...". Wait! You already used $x=7$, don't you have to use a different $x$ in the second statement?... The answer is, no, I don't have to. I can choose any $x$ so I can choose the same $x$ in both statements. $\endgroup$ – fleablood Apr 3 '17 at 18:59
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Let me redo the definition and proof to emphasize points that are implicit in your text but elided over and causing you confusion (reasonable confusion). The pertinent emphasis is in itatics and note the subscript of the deltas:

Definition $\lim_{x\rightarrow a} f(x) = L$ if for any possible $\epsilon > 0$ at all there will exist a $\delta_{f,\epsilon} > 0$ that is dependent upon both the function $f$ and the choice of $\epsilon$ so that whenever we have $|x - a| < \delta_{f,\epsilon}$ it will always follow that $|f(x) - L | < \epsilon$.

Squeeze Theorem: We have $\lim_{x\rightarrow a} f(x) = L$ and $\lim_{x\rightarrow a} g(x) = L$ and $g(x) \le h(x) \le f(x)$ for all $x$ then $\lim_{x\rightarrow a} h(x) = L$.

Proof: We need to show that for any arbitrary $\epsilon > 0$ we can choose we will be able to find a $\delta_{h,\epsilon} > 0$ that is dependent on the function $h$ and our choice of $\epsilon$ so that $|x-a|< \delta_{h,\epsilon}$ will imply $|h(x) - L| < \epsilon$.

For any arbitrary $\epsilon$ we know that a $\delta_{f,\epsilon}$ exists that will ... do the stuff... for $f$ and for $\epsilon$. And we know that for the same $\epsilon$ we can find a different $\delta_{g,\epsilon}$ that will ... do the stuff ... for $g$ and for the same $\epsilon$. (Because such a $\delta$ will exist for every positive value, it will have to exist for the same $\epsilon$.)

We claim that if $\delta_{h,\epsilon} = \min (\delta_{g,\epsilon},\delta_{g ,\epsilon})$ then we will be able to show that $\delta_{h,\epsilon}$ will do the stuff for the function $h$ and the same arbitrary value of $\epsilon$.

.... and then we do the proof.

==== old answer ====

No.

The point is the function $f,g$ to have limits the condition must be true for ALL possible $\epsilon$s. They will both be true for all $\epsilon$s So there is no need at all to use different $\epsilon$s and we want to use the same one to get a universal result.

In fact if we had to use a specific $\epsilon$ (rather than any arbitrary $\epsilon$) to get the result, that would mean the function does not have that limit at that point.

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  • $\begingroup$ ok, So the relationship between $\delta$ and $\epsilon$ on all the functions may be different, but the $\epsilon$ is the same in all of them and what changes is how you define the $\delta$ for each function. but since all i need is for $|x-a|<\delta$ then i can use $min(\delta_1,\delta_2)$. is that right? $\endgroup$ – Joaquin Brandan Apr 3 '17 at 19:12
  • $\begingroup$ for more clarification, since all i need is for $|x−a|<\delta_1$ and $|x−a|<\delta_2$ then it is also correct to use $|x−a|<min(\delta_1,\delta_2)$ $\endgroup$ – Joaquin Brandan Apr 3 '17 at 19:30
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    $\begingroup$ You currently have two deltas if |x - a| < d1 then |f(x)-L| < e. If |x-a|<d2 then |g(x)-L|< e. If |x-a|< min(d1,d2) then |f(x)-L| < e and |g(x)-L| < e. So -e < g(x) - L < h(x)-L < f(x)-L < e so |h(x) -L| < e. So setting d3 to min(d1,d2) does the trick. $\endgroup$ – fleablood Apr 3 '17 at 19:31
  • $\begingroup$ Yes, if $k < \min(a,b)$ then both $k < a$ and $k < b$ are both true statements. $\endgroup$ – fleablood Apr 3 '17 at 19:32
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    $\begingroup$ The thing is $|a-b| < d1$ is a conditional statement $|a-b|< d2$ one or the other or both or neither is true. But we can find a tighter condition that if $|a-b| < min(d1,d2)$ then they both must be true. $\endgroup$ – fleablood Apr 3 '17 at 19:34
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You can choose the same $\epsilon_0$ because $(1),(2),(3)$ are true $\forall \epsilon \in \Bbb R, \epsilon\gt0$.

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Recall that $\lim_{x \to c} f(x) = L$ means: $$\forall \varepsilon > 0 \, \exists \delta > 0 \, \forall x \in \operatorname{dom}(f) \, (0 < |x-c| < \delta \implies |f(x) - L| < \varepsilon).$$ This means that if you're giving an $\varepsilon$-$\delta$ proof, the structure of your proof should be like so:

"Let $\varepsilon > 0$."
      [A choice for $\delta$ goes here.]
            "Let $x \in \operatorname{dom}(f)$."
                  "Suppose $0 < |x-c| < \delta$."
                  [Proof that $|f(x) - L| < \varepsilon$ goes here.]
"Thus, $\lim_{x \to c} f(x) = L$."

So from the very beginning of your proof, you've already taken an arbitrary $\varepsilon > 0$. That's why you haven't got several different epsilons floating around.

By the way, why does one take $\min \{ \delta_1, \delta_2 \}$ for $\delta$? You do it so that when you assume $0 < |x-c| < \delta$, you immediately get both $0 < |x-c| < \delta_1$ and $0 < |x-c| < \delta_2$, which then implies both $|f(x) - L| < \square$ and $|g(x) - L| < \boxtimes$, whatever $\square$ and $\boxtimes$ may be. Then you can go from there.

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  • $\begingroup$ isnt each $\epsilon$ written in terms of $\delta$ and affected by it's function?, If i prove an arbitrary limit then i get a specific relation between delta and epsilon, i understand that the specific relation is not important as long as it exists, but the relation on 1) could be diferent from the relation on 2. does that make any sense? $\endgroup$ – Joaquin Brandan Apr 3 '17 at 18:30
  • $\begingroup$ No, because notice the order of $\varepsilon$ and $\delta$ in the definition: $\varepsilon$ comes first, and $\delta$ next, so $\delta$ depends on $\varepsilon$. That's why after you take an arbitrary $\varepsilon > 0$, you have to work quite hard to find a $\delta > 0$ that will do the trick. $\endgroup$ – Mark Twain Apr 3 '17 at 18:41
  • $\begingroup$ By the way, how do you choose $\delta$? This is the most difficult part. You need to focus on the statement $|f(x) - L| < \varepsilon$. That's the last statement you're showing anyway. So focus on it, and you'll get to a point where a good choice for $\delta$ shows up. This is part of your scratch work. $\endgroup$ – Mark Twain Apr 3 '17 at 18:48

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