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Possible duplicate of: Epsilon delta proof min

see: https://math.libretexts.org/Bookshelves/Calculus/Book:_Calculus_(Apex)/01:_Limits/1.02:_Epsilon-Delta_Definition_of_a_Limit where the following example is featured.

Prove: $$\lim\limits_{x \to 4} \sqrt{x} = 2$$

\begin{align*} 2 -\varepsilon &< \sqrt{x} < 2 + \varepsilon \\ (2 -\varepsilon)^2 &< x < (2 + \varepsilon)^2 \\ 4- 4\varepsilon +\varepsilon^2 &< x < 4 + 4\varepsilon + \epsilon^2 \\ 4- (4\varepsilon -\varepsilon^2) &< x < 4 + (4\varepsilon + \varepsilon^2)\\ \end{align*}

Here $\delta$ has two possible values $4\varepsilon - \varepsilon^2$ and $4\varepsilon + \varepsilon^2$. Both values lead to the conclusion that $\lim\limits_{x \to 4} \sqrt{x} = 2$, so which should be used?(Shown to be an incorrect statement in the answers)

Arguments for $\delta \leq 4\varepsilon + \varepsilon^2$.

$\delta > 0 \forall \varepsilon$ = delta is positive for all epsilon

$\delta \leq 4\varepsilon - \varepsilon^2 < 4\varepsilon + \varepsilon^2$ i.e. it is larger than $4\varepsilon - \varepsilon^2$ so delta covers more values.

Arguments for $\delta \leq 4\varepsilon - \varepsilon^2$

As proven by Limits with epsilon-delta - the accepted answer, it doesn't matter if \epsilon has an upper bound (in this case, $\varepsilon \lt 4$) because:

  1. epsilon should be small
  2. if there exist a delta for $\varepsilon \in (0,t)$ the same delta works for $\varepsilon \geq t$

Why is the minimum used?

2 Answers2

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When you choose a value of $\delta$ corresponding to any particular $\varepsilon,$ you are asserting that $2 - \varepsilon < \sqrt x < 2 + \varepsilon$ whenever $4 - \delta < x < 4 + \delta.$

Let's try a concrete example: what happens if $\varepsilon = 0.1$?

If you say that $\delta = 4\varepsilon + \varepsilon^2,$ then you are saying that you can set $\delta = 4\times 0.1 + 0.1^2 = 0.41$ and then it will be true that $1.9 = 2 - 0.1 < \sqrt x < 2 + 0.1 = 2.1$ whenever $3.59 = 4 - 0.41 < x < 4 + 0.41 = 4.41.$

But what if $x = 3.591025$? Then $3.59 < x < 4.41,$ so you have satisfied the "whenever $4 - \delta < x < 4 + \delta$" condition, but $\sqrt x = 1.895,$ so it is not true that $1.9 < \sqrt x < 2.1$

In short, the formula $\delta = 4\varepsilon + \varepsilon^2$ does not work for this particular value of $\varepsilon.$ If you look further into this you should be able to show that the formula does not work for any other values of $\varepsilon$ either.

The thing behind all this is that in a delta-epsilon proof, we only assert the existence of one value of $\delta$ for any particular value of $\epsilon,$ and the same value of $\delta$ has to work in both directions, both below and above the limiting value of $x.$

However, we never said that we have a $\delta$ that gives all values of $x$ for which $L - \varepsilon < f(x) < L + \varepsilon.$ In your proof you don't need to show that $2 - \varepsilon < \sqrt x < 2 + \varepsilon$ if and only if $4 - \delta < x < 4 + \delta$; you only need to show the "if" direction.

And this leads into an observation about delta-epsilon proofs in general, which you may want to repeat as a mantra until you have fully internalized it:

You can never go wrong by choosing $\delta$ "too small", as long as you keep it positive.

This is how we can make use of a definition that requires us to use the same $\delta$ in both directions: even though the complete interval of values of $x$ that satisfy $L - \varepsilon < f(x) < L + \varepsilon$ may be asymmetric, we only need to identify a subset of that interval, and it is always possible to find a symmetric subset of an asymmetric interval around a particular value of $x.$

So you can never go wrong by taking the smaller of two positive values. If the interval of $x$ values is asymmetric, the distance to the farther end of the interval is irrelevant. For that matter, you don't even need to be sure what the exact distance is to the nearer end of the interval. You just need to be sure that whatever that distance is, the $\delta$ you choose is not larger than that distance. Smaller is fine.

On the other hand you will always go wrong if you choose $\delta$ too large.

David K
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  • ''However, we never said that we have a$\delta$ that gives all values of $x$ for which $L−\epsilon<f(x)<L+\epsilon$.'' See question:https://math.stackexchange.com/questions/3734353/epsilon-delta-on-a-function-with-restricted-range where it says in an answer : for all for all $x \in (c-\delta,c+\delta)\ \cap \ \mathrm{dom}_{f} $ there exist a $y \in (L-\varepsilon,L+\varepsilon)$ such that $y=f(x)$. –  Jul 25 '20 at 18:10
  • What I said is the same as what the other answer said. What the other answer does not say is that for every $y \in (L-\varepsilon,L+\varepsilon)$ there exists an $x \in (c-\delta,c+\delta) \cap \mathrm{dom}_f$ such that $y = f(x)$. That is, it never says there is a $\delta$ that gives all values of $x$ such that $f(x) \in (L-\varepsilon,L+\varepsilon)$. – David K Jul 25 '20 at 18:16
  • Also quoting from the answer you quoted, "The definition requires $(a - \delta, a + \delta)$ to map into $(L - \varepsilon, L + \varepsilon)$, not onto it." If we said we had a $\delta$ that gives all values of $x$ for which $L - \varepsilon < f(x) < L + \varepsilon,$ we would be claiming that $(a - \delta, a + \delta)$ mapped onto $(L - \varepsilon, L + \varepsilon)$, not just into it. Which would be a bad claim to make, because it is not always possible to find such a $\delta.$ – David K Jul 25 '20 at 18:20
  • Doesn't into mean an injection- where all $x \in X$ are mapped onto $y \in Y$, where in this case $Y = (L-\varepsilon,L+\varepsilon)$ and $X = (c-\delta,c+\delta)$ –  Jul 26 '20 at 08:46
  • $4 - (4\varepsilon - \varepsilon^2) < x < 4 + (4\varepsilon +\varepsilon^2)$ i.e. $ x \in (4 - (4\varepsilon - \varepsilon^2),4 + (4\varepsilon +\varepsilon^2))$ however $(4 - (4\varepsilon - \varepsilon^2),4 + (4\varepsilon +\varepsilon^2)) \subset ( 4 -(4\varepsilon +\varepsilon^2,4 + (4\varepsilon +\varepsilon^2))$ so if we define $\delta \leq 4\varepsilon +\varepsilon^2$ then we say $x \in ( 4 -(4\varepsilon +\varepsilon^2),4 + (4\varepsilon +\varepsilon^2))$ but $ x \notin ( 4 -(4\varepsilon +\varepsilon^2),4 - (4\varepsilon -\varepsilon^2))$ –  Jul 26 '20 at 09:10
  • which is why $x= 3.591025$ didn't map into $(L-\varepsilon,L+\varepsilon)$ and $\delta$ is not $4\varepsilon + \varepsilon^2$ i.e. a proof by contradiction using the statement: for all $x \in (c−\delta,c+\delta ) \cap \mathrm{dom}_f$ there exist a $y\in(L−\varepsilon,L+\varepsilon)$ such that $y=f(x)$. so why is this statement wrong. –  Jul 26 '20 at 09:10
  • What does ''too large'' and ''too small'' mean - don't epsilon-delta proofs have to be precise/formal. –  Jul 26 '20 at 09:13
  • https://www.quora.com/What-is-the-difference-between-mapping-into-versus-mapping-onto#:~:text=%E2%80%9CInto%E2%80%9D%20and%20%E2%80%9Conto%E2%80%9D,throughout%20mathematics%20in%20this%20way.&text=The%20function%20is%20a%20surjection,least%20one%20in%20such%20that%20.. says that ''into'' means injection which would suggest that $\forall x_1,x_2 \in (c−\delta,c+\delta)\setminus {c} \cap \mathrm{dom}_f$ if $f(x_1)=f(x_2)$ then $x_1=x_2$ . –  Jul 29 '20 at 12:24
  • however this does not make sense say for f(x)=1, I choose 2 distinct points $a_1,a_2\in (c−\delta,c+\delta)\setminus {c} \cap \mathrm{dom}_f$ then $f(a_1)=f(a_2)=1$ however $a_1 \neq a_2$ by definition thus the function is not injective but it is still (obviously) continuous so shouldn't it be a map onto (surjection) not into (injection). –  Jul 29 '20 at 12:24
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    When the quoted answer said "map into", it meant merely that the image of $(c−\delta,c+\delta)\setminus {c} \cap \mathrm{dom}_f$ under the mapping $x \mapsto f(x)$ was a subset of $(L - \varepsilon, L + \varepsilon).$ It did not mean that the mapping was one-to-one. The exact term for a one-to-one map is "injection." I know there are some sources that call a one-to-one map an "into-map", but that's not how mathematicians usually describe such functions and it's not what they usually mean by "into." – David K Jul 29 '20 at 12:38
  • isn't that the equivalent of my statement in the first comment where the set of all $y's$ for all $x$ form the image of $f$ in the interval $(L-\epsilon,L+\epsilon)$ i.e. $\forall x \in (c-\delta,c+\delta)\setminus { c } \cap \mathrm{dom}_f$ there exist no $f(x)$ that is outside $(L-\epsilon,L+\epsilon)$ –  Jul 29 '20 at 12:49
  • Yes, they are the same. The point is merely that you have to think about what that statement implies and what it doesn't imply. It doesn't imply that the values you map from $(c−\delta,c+\delta)\setminus{c}\cap\mathrm{dom}_f$ to $(L-\varepsilon,L+\varepsilon)$ have to "fill up" $(L-\varepsilon,L+\varepsilon).$ They don't even have to be be a "good fit" for $(L-\varepsilon,L+\varepsilon).$ They could map to just a tiny fraction of the interval. But if one single value from the input interval maps to a value outside $(L - \varepsilon, L + \varepsilon),$ you have a bad $\delta.$ – David K Jul 29 '20 at 13:03
  • You do have to prove 'iff' not just 'if' since 'iff' works in both directions, if it was an 'if' then it would destroy the derivations of e-d proofs https://en.wikipedia.org/wiki/If_and_only_if#Distinction%20from%20%22if%22%20and%20%22only%20if%22 –  Aug 24 '20 at 19:18
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    When you have a theorem of the form "$P$ iff $Q$" then you have to prove both directions. But the epsilon-delta definition of limits has only an if statement, not iff. – David K Aug 24 '20 at 20:03
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"Both values lead to the conclusion that $\lim\limits_{x \to 4} \sqrt{x} = 2$" is not true.

What you need is a $\delta$ such that for all $x\in[4-\delta,4+\delta]$ the condition holds, so you need

$$[4-\delta,4+\delta]\subseteq[4- (4\varepsilon -\varepsilon^2),4 + (4\varepsilon + \varepsilon^2)].$$

This requires

$$\delta\le4\varepsilon -\varepsilon^2, 4\varepsilon + \varepsilon^2.$$