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:
- epsilon should be small
- if there exist a delta for $\varepsilon \in (0,t)$ the same delta works for $\varepsilon \geq t$
Why is the minimum used?
Best Answer
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:
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.