Explanation for the formal limit definition

Question

I have been trying to understand the formal definition of limit, including MSE posts, other universities' notes, my math textbook, etc. I can't seem to understand its usage well. I get that if for every interval around x we can find an f(x) within an interval around f(x). I also sort of understand that if I make the $|f(x)-L|<\epsilon$ into a similar form as $|x-a|<\delta$ I can get $\delta(\epsilon)$. However, I am completely lost at how to prove that the guessed $L$ is correct. It looks as if I am simply substituting $\delta(\epsilon)$ into $|f(x)-L|<\epsilon$, which seems like I am repeating the same step backwards? How does that contribute to the proof?
I apologise for the simple question, but I can't seem to get past this. Could someone explain it to me in simple 5-year old English how the definition is used at every step of the way? Thank you very much for your help.
Edit: typo I apologise for the lack of a specific question, as user2661923 has kindly suggested. I have 2 right now. I am actually trying to prove an infinite limit, which was thrown to me before my lecturer has taught the formal definition. Thus, I had to go look at formal definition questions where the limit does exist. This is the infinite limit question:
$$\lim_{x\to 2^+}[\frac{1}{2-x}-\frac{3}{8-x^3}]$$
I managed to do this and then got stuck:
$$\lim_{x\to 2^+}\frac{(x+1)^2}{8-x^3}$$ So I went to formal definition questions:
Prove that $$\lim_{x\to 3}(4x-5)=7$$
I understand up to: $$0<|x-3|<\delta$$ $$0<|(4x-5)-7|<\epsilon$$ $$|x-3|<\frac{\epsilon}{4}$$ And thus $$\delta(\epsilon)=\frac{\epsilon}{4}$$ But then I got stuck trying to figure out how to prove that the L satisfies these conditions.
Once again, thank you very much for your help and suggestions.

Answer 1

To prove a limit $L$, you need to solve the following logical equation:

$$\forall\epsilon>0:\exists \delta>0:\forall x:0<|x-x_0|<\delta \implies |f(x)-L|<\epsilon.$$

In the case that $f(x)$ is invertible in some interval around $x_0$, we can write

$$L-\epsilon<f(x)<L+\epsilon$$ and assuming a growing function,

$$f^{-1}(L-\epsilon)<x<f^{-1}(L+\epsilon)$$

or

$$f^{-1}(L-\epsilon)-x_0<x-x_0<f^{-1}(L+\epsilon)-x_0.$$

From this, it should be clear that any $\delta$ such that

$$\delta\le\min(x_0-f^{-1}(L-\epsilon), f^{-1}(L+\epsilon)-x_0)$$ can do.

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With your example, $f(x)=4x-5$, and $f^{-1}(y)=\dfrac{y+5}4$. Hence

$$\delta\le\min\left(3-\dfrac{7-\epsilon+5}4,\dfrac{7+\epsilon+5}4-3\right)=\frac\epsilon4$$ will work.

As you can check,

$$\forall\epsilon>0:\forall x:0<|x-3|<\frac\epsilon4\implies |4x-12|<\epsilon$$ as expected.

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In the general case, the inequation

$$|f(x)-L|<\epsilon$$ may have solutions in $x$, and $\delta$ must be such that $(x_0-\delta,x_0+\delta)$ is wholly contained in the solution set, if possible. It is your task to solve the inequation. An open subinterval of the solution set can do, provided it contains $x_0$.

Answer 2

It looks as if I am simply substituting $\delta(\epsilon)$ into $|f(x)-L|<\epsilon$, which seems like I am repeating the same step backwards?

That is part of the actual proof. The calculations you make before to find a suitable $\delta$ is prework that does not need to be presented.

When writing a limits proof, I find it useful to start with $|f(x)-L|$ and try to factor out $|x-a|,$ which is later replaced with $\delta$. The factor $|x-a|$ is then replaced with $\delta,$ and I need to make sure that the other factor doesn't grow too fast, and find $\delta$ small enough so that the less-than-chain can end with $<\epsilon$.

Example

Show that $\lim_{x\to 3} (4x-5) = 7$.

Prework $$ |(4x-5)-7| = |4x-12| = 4|x-3| < 4\delta $$ Here it's clear that by taking $\delta<\frac14 \epsilon$ we can continue with $$ < 4 \cdot \frac14 \epsilon = \epsilon . $$

How the proof is written

Given $\epsilon>0$ let $\delta=\frac14\epsilon.$ Then, when $|x-3|<\delta$, we have $$ |(4x-5)-7| = |4x-12| = 4|x-3| < 4\delta = 4 \cdot \frac14 \epsilon = \epsilon. $$