When proving $\lim_{x\to a} x^2 = a^2$ we need to show that for all $\epsilon > 0$ there is some $\delta > 0$ so that if $0<|x-a|<\delta$ then $|x^2-a^2|< \epsilon$. Now writing $|x^2-a^2|$ as $|x-a||x+a|$ why can't we just pick $\delta = \frac{\epsilon}{|x+a|}$. Why is the reason that $\delta$ must be independent of $x$?
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- Because this is how limits are defined. 2. Because the phrase $$\forall\epsilon\ \exists\delta(x)\ \forall x\ |x-x_0|<\delta(x)\implies |f(x)-\ell|<\epsilon$$ introducing $\delta(x)$ before $x$, makes no sense. 3. Because every function defined at $x_0$ fulfills the condition $$\forall\epsilon\ \forall x\ \exists\delta(x)\ |x-x_0|<\delta(x)\implies |f(x)-f(x_0)|<\epsilon$$ (Item 3. is slightly subtler than 1. and 2.)
– Did Mar 11 '18 at 08:47 -
@spkakkar Hmmm... no. – Did Mar 11 '18 at 08:51
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@spkakkar how exactly is this a duplicate. These questions are asking two different things. – user500668 Mar 11 '18 at 08:55
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You are right. I was too early to comment :P Sorry. I am deleting it. – spkakkar Mar 11 '18 at 08:59
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Well it is the other way round. Your $x$ depends on $\delta$. Based on $\epsilon$ some $\delta $ is chosen such that all $x$ lying in the range $(a-\delta, a+\delta) $ fulfill some specific condition. So the $\delta$ gives you the range of $x$ for the which some condition needs to be verified. – Paramanand Singh Mar 11 '18 at 17:41
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Moreover the way you are trying to find $\delta$ looks more like algebraic manipulation of inequalities. Proceeding in this manner is equivalent to missing the whole point of limit concept. By definition of limit if some $\delta$ works for a given $\epsilon $ then any smaller $\delta$ also works for that $\epsilon$. Thus one is not supposed to solve for $\delta $ in terms of $\epsilon $ like solving some algebraic equation.. – Paramanand Singh Mar 11 '18 at 17:47
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The definition of limit states:
$$\lim_{x \to a}f(x) = L :\iff\forall \epsilon > 0: \exists \delta > 0: \forall x \in dom(f) : (0 < |x-a| < \delta \implies|f(x)-L| < \epsilon)$$
From a logic point of view, $\delta$ can only depend on the things that are introduced before the $\delta$ is introduced. I.e., here $\delta$ can only depend on $\epsilon$. This is a rule that generally works for expressions with nested quantifiers.
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If you want a more intuitive explanation: $\delta$ is the size of a range of $x$ values, and it doesn't make any sense for the size of an interval to depend on the values in the interval. To make a real-world analogy, it would be as if the length of a road depended on how far along the road you are.
David Z
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Another attempt:
$x \in D_f$, real.
$\lim_{x \rightarrow a} f(x) = L$
$ \iff$
Let $\epsilon >0$ be given (you are free to choose),
then there exists a $\delta (a,\epsilon) >0$, such that
$|x-a| \lt \delta$ implies $|f(x)-L| \lt \epsilon.$
The order:
1) $a$, real is given.
2) Choose any $\epsilon$.
3) If a $\delta$ with the above mentioned
properties exists then $\delta$ depends on $a$ and $\epsilon.$
4) Now restrict your $x$ to:
$|x-a| \lt \delta(a,\epsilon)$,
then: $|f(x)-L| \lt \epsilon.$
Peter Szilas
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You are trying to find a single $\delta$ value ahead of time that works for many cases. The cases are where $0 < |x-a| < \delta$, and the definition involves trying to find one delta $\delta$ that works for all such cases. Since there are many cases that you have to account for, each case having its own $x$ value, you cannot in any way depend on a particular $x$ value.
However, if you could find out any information that applies to all $x$ values, then you could absolutely use that in your choice for delta. For example, if there was a fact that $x < 10$ for all $x$, then you could use that information to come up with a $\delta$. However, for the $\epsilon-\delta$ definition, we are not given any additional constraints on $x$ before $\delta$ comes into existence, so I do not believe there are any facts you could gather generally about $x$ to help you here. You have to bring $\delta$ into existence before you can learn anything about $x$, and therefore $\delta$ cannot depend on $x$.
So, the only thing you can use to help you choose $\delta$ is that for any $x$ we must have $0 < |x - a| < \delta \implies |f(x) - L| < \epsilon$
Hope that makes sense :)
David Callanan
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