$\displaystyle \lim_{x\to 1} \frac{1}{x-1} $ does not exist.
Proof
The definition for limit doesn't exist is
$\forall L\in \mathbb R,\exists \epsilon>0 ,\forall \delta>0 ,\exists x\in domf$ such that $|x-1|<\delta$ and $|\frac{1}{x-1}-L|\geq \epsilon$
Chose $x=\frac{1}{n}+1>0,n\in \mathbb N$. Then based on Archimedean property $|x-1|=|\frac{1}{n}+1-1|=\frac{1}{n}<\delta$
Also $|\frac{1}{x-1}-L|\geq \epsilon$ means that $\frac{1}{x-1}\geq L+\epsilon$ or $\frac{1}{x-1}\leq L-\epsilon$
Because $x=\frac{1}{n}+1>0$, $\frac{1}{x-1}=n$. Also based on Archimedean property, there exists a natural number such that it is greater than $L+\epsilon$
Then choose $epsilon$ to be any positive numbers.
Hence, $\displaystyle \lim_{x\to 1} \frac{1}{x-1} $ does not exist.
Is this correct? Also, I'm kind of confused about the quantifiers in the limit statement.
In the sentence "$\forall L\in \mathbb R,\exists \epsilon>0,\forall \delta>0,\exists x\in domf$, does this mean that there exists a constant epsilon, just like "1,2,…".Then this fixed constant should satisfy all delta. And for every delta, there is a $x$. In the end, epsilon is kind of “isolated”，but x is "accompanying" with the delta.
Best Answer
You have all the right ingredients and all the right tools, but I can see from your write-up alone that you are finding the four nested quantifiers difficult. And it is difficult to properly wrap your head around. It might help if you start with a skeleton for your proof, that you can fill in.
Because your statement starts with $\forall L \in \Bbb{R}$, a good sentence to begin your proof would be
The next quantifier is $\exists \varepsilon > 0$. So, at some point after this first sentence, we will need
The specific value for this $\varepsilon$ is for us to decide. This should be based on our scratch work. The next quantifier is $\forall \delta > 0$, so once again, we will need the following:
Based on our final quantifier, we then want:
That's when we can start proving the predicate at the end of the quantifiers: that $x \in \operatorname{dom} f$, that $|x - 1| < \delta$, and that $|\frac{1}{x - 1} - L| \ge \varepsilon$.
Now, let's fill in the gaps. From your proof, $\varepsilon > 0$ seemed to be arbitrary. All you needed was some $n \ge L + \varepsilon$. So, let's just pick $\varepsilon = 1$. And why not! You should be able to find some $n \ge L + 1$.
Next, we need to figure out our $x$. You chose $x = \frac{1}{n} + 1$, which is potentially sensible, but you should really define your $n$ first. How was $n$ defined in your proof? It was defined to satisfy two properties:
Both of these are sensible conditions based on the Archimedean principle. Let's define: $$n = \max\left\{\lceil L \rceil + 1, \left\lceil \frac{1}{\delta}\right\rceil + 1\right\},$$ where $\lceil \cdot \rceil$ is the ceiling function (which only exists due to the Archimedean property!). Then, it makes sense to define $x = 1 + \frac{1}{n}$. With this in mind, I would write the proof as follows: