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.

  • 3
    $\begingroup$ I upvoted, because a lot of effort went into your posting. Further, although it is unclear what your asking, to a large extent, your lack of clarity is attributable to the concepts being difficult to intuit. However, I have to hold off giving an answer (as others may also feel), because I'm not sure "hot to hit the nail on the head." It would help, if you could edit your question with a specific problem that (perhaps) you've already wrestled with. The idea is that having a mathSE reviewer solve the specific math problem that you pose should "hit the nail on the head" for you. $\endgroup$ Sep 1 '20 at 6:58

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,




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.

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.

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 $x_0$.

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  • $\begingroup$ I am not addressing the question with the infinite limit to keep it simple. $\endgroup$
    – user65203
    Sep 1 '20 at 8:07

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 $<\epsilon$.


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 $$ |(4x-5)-7| = |4x-12| = 4|x-3| < 4\delta = 4 \cdot \frac14 \epsilon = \epsilon. $$


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