Definition of the Limit: Question about $|f(x)-L|$ manipulation to find a suitable delta

Question

In the definition of the limit, often the first step to find a suitable delta is to manipulate $|f(x)-L|$ to look something like $|x-a|$, after that we can see delta in terms of epsilon that we can choose as a guess.

I would like to ask why is it that we can use this manipulation as delta and end up satisfying the statement:

$|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$

All the videos I have watched, after extracting $|x-a|$ from $|f(x)-L|$ goes like 'look! we have found an appropriate delta, now we have to check if this works'. I am having trouble understanding how this indeed satisfies the condition, or how the extraction provides us with a good guess for delta.

Say for example, given $\lim_{x\to 1}(2x+3)=5$

We write $|2x+3-5|$, and then manipulate that to be: $|2x-2|$, $2|x-1|$. We found $|x-a|$ (in this case, $|x-1|$), so we say that the suitable $\delta$ is $\epsilon/2$, because we manipulated $|2x-2|<\epsilon$ to be $|x-1|<\epsilon/2$

How does this manipulation give us a good guess for delta?

Answer 1

In the example $$\lim_{x\to 1}(2x+3)=5$$ you have $f(x) = 2x+3$ and $L=5,$ so as you showed, you write $$\lvert f(x)-L \rvert = \lvert (2x+3)-5 \rvert = \lvert 2x - 2\rvert. $$

You also have $a = 1,$ so $\lvert x - a \rvert = \lvert x - 1 \rvert.$ So if you notice that $\lvert 2x - 2\rvert = 2\lvert x - 1\rvert,$ you will have discovered that $$\lvert f(x)-L \rvert = 2\lvert x - a \rvert. $$

Since the left and right sides of this equation are equal quantities, we know that $\lvert f(x)-L \rvert < \epsilon$ if and only if $2\lvert x - a \rvert < \epsilon,$ while $2\lvert x - a \rvert < \epsilon$ is true if and only if $\lvert x - a \rvert < \frac\epsilon2$.

Since each of these inequalities is true if and only if the others are true, we can work backwards from the end:

If $\lvert x - a \rvert < \frac\epsilon2$ then $2\lvert x - a \rvert < \epsilon.$

If $2\lvert x - a \rvert < \epsilon$ then $\lvert f(x)-L \rvert < \epsilon.$

Therefore, if $\lvert x - a \rvert < \frac\epsilon2$ then $\lvert f(x)-L \rvert < \epsilon.$

That is the implication in the definition of $\lim_{x\to a}f(x) = L,$ except that we have $\frac\epsilon2$ where the definition has $\delta.$

So for any $\epsilon > 0,$ we know that there is a number that we can call $\delta$ and that happens to be equal to $\frac\epsilon2$ such that $\lvert f(x)-L \rvert < \epsilon$ whenever $\lvert x - a \rvert < \delta.$ And that is exactly the definition of the limit.

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I think it is important to remember that the step of "finding $\delta$ in terms of $\epsilon$" is a technique to help you prepare to write a delta-epsilon proof; it is not part of the proof itself. Indeed, if you were very lucky and guessed (without doing any calculations at all) that you could set $\delta$ equal to $\frac\epsilon2$ in your proof, you would find when you wrote the proof that this actually worked as desired. But most of us are not such lucky guessers, so we need a little calculation to help us with the guesses.

But there is no general method (at least none I am aware of) to tell you how to manipulate any arbitrary differentiable function $f(x)$ in order to find the derivative of $f(x)$ at any particular value of $x.$ You might be able to generalize the method above to work for any linear function, but you will need a different method for a quadratic function, another for an exponential function, etc.

So even with these techniques, the answer to "how did you know" is generally that you did not know how to do it when you started, but you used intuition, experience, and sometimes trial and error (the manipulation looked like it would get you a suitable way to express $\delta$, but it did not) until you found a proof that worked.

Answer 2

$|f(x)-L| < \epsilon$ whenever $|x-a|<\delta$ satisface because the process you construct your $\delta$.
The process of manipulate $|f(x)-L|$ is precisely to garante that when you do the inverse process, the obtained $\epsilon$ is small enough
Eg suppouse we wanna show that
$f(x)=x²+7x+6$ is continues at $x=1$
First, note that we need that $|x²+7x+6-14|<\epsilon$ but this is
$|x²+7x-8|<\epsilon$
On the other hand, $x²+7x-8$ can be factorized as $(x+8)(x-1)$.
We want to show that, no matter how small is the distance ($\delta$) between a point $x$ and $1$ to form an open interval in the domain, it always exist an open interval in range that contains $f(1)$ so, start by assuming a distance $\delta=1$ so
$|x-1|<1$ so if this happens, I need to know what happens with $|x+8|$? Note that
$|x+8|=|x-1+1+8|<|x-1|+9<1+9=10$ this last, by our first assumption, so this means that it will be helpful using $\delta= \frac{\epsilon}{10}$
So we can start by saying,
let be $\delta=min(1,\frac{\epsilon}{10})$ and so on... this election of $\delta$, guarantees that
$|x+8||x-1|$ will be always minor than $\epsilon$
I hope this can help you bro.