some quesitons about to prove$\lim_{x\to5} \frac{1}{x-1}=\frac{1}{4}$ Medthod 1 Prove $$\lim_{x\to5} \frac{1}{x-1}=\frac{1}{4}$$
According to the definition of limit, $\mid x-5\mid<\delta\implies \mid \frac{1}{x-1}-\frac{1}{4}\mid<\epsilon$.
Let $$\mid \frac{1}{x-1}-\frac{1}{4}\mid<\epsilon\\
\frac{1}{4}\mid\frac{4-x+1}{x-1}\mid<\epsilon\\
\mid 5-x\mid<4\epsilon\mid x-1\mid
$$
As $x$ is approaching $5$ so that it is safe to assume$\mid x-5\mid<1$. Then we have $$\mid x-5\mid<1\implies 4<x<6\implies 3<x-1<5$$.
Then $\mid 5-x\mid<4\epsilon\mid x-1\mid<12\epsilon$. Then let $\delta\leq min\{1,12\epsilon\}$ and work it backward to obtained the results $\mid \frac{1}{x-1}-\frac{1}{4}\mid<\epsilon$ as desired.
Q1 Is that always safe to assume $x$ approaching to the constant provided in the similar question within the radius of 1?
Method 2.
Again, let $$\mid \frac{1}{x-1}-\frac{1}{4}\mid<\epsilon\\ -\epsilon<\frac{1}{x-1}-\frac{1}{4}<\epsilon\\1+\frac{4}{1+4\epsilon}<x<1+\frac{4}{1-4\epsilon}$$
Besides that, $$\mid x-5\mid<\delta\implies 5-\delta<x<5+\delta$$ Then we could set $$1+\frac{4}{1+4\epsilon}\leq5-\delta<x<5+\delta\leq 1+\frac{4}{1-4\epsilon}$$
Q2 I do not understand this one, why does the results obtained from $\delta$ will be bounded by that of $\epsilon$
Then we have
$$\left\{ \begin{array}{ll} \delta\leq\frac{4}{1-4\epsilon}-4 \quad \\ \delta\leq 4-\frac{4}{1+4\epsilon} \end{array} \right.$$ we take $\delta=min\{\frac{4}{1-4\epsilon}-4，4-\frac{4}{1+4\epsilon}\}$ However, so far in method 2 is just scratch work, and I cannot work it back from here to get $\mid \frac{1}{x-1}-\frac{1}{4}\mid<\epsilon$ as desired
Appreciated it for anyone could answer my Q1&Q2, and show me how to work it back for method 2
 A: The idea is to find a specification between $\delta$ and $\epsilon$ so that for any $\epsilon > 0$
if $0 < |x - 5| < \delta$ 
then $\displaystyle \left|~\frac{1}{x-1} - \frac{1}{4} ~\right| < \epsilon.$
Unless I missed something, your work didn't seem to take the approach of saying:

*

*Here is the specification of $\delta$ against $\epsilon$.


*Based on this specification, 
$\displaystyle 0 < |x - 5| < \delta
 \implies \left|~\frac{1}{x-1} - \frac{1}{4} ~\right| < \epsilon.$
I am supposed to focus on your work, and help you take your work to completion.  However, your approach is totally different than mine.  I therefore ask that you consider my approach, and then post questions, following my answer, as you need to.
I will attack the problem by first deriving a specification, and then verifying that the specification works.
My first step is to simplify the fraction:
$\displaystyle \left|~\frac{1}{x-1} - \frac{1}{4} ~\right|
= \left|~\frac{5-x}{4(x-1)} ~\right|.$
Next, I will start with the constraints on $\delta$ and $x$.
$x \neq 5$ and 
$|x - 5| < \delta \implies - \delta < (x-5) < \delta
\implies 5-\delta < x < 5 + \delta$.
Next, I'm going to analyze the fraction
$$\left|~\frac{5-x}{4(x-1)}~\right|.\tag{1}$$
My thinking is that when $x$ is very close to $5$, I want to establish an upper bound on the numerator in expression (1) above, and a lower bound on the denominator.  This type of approach will imply an upper bound on the fraction.
I am going to add an artificial contrivance that in addition to a relationship between $\delta$ and $\epsilon$, I will also insist that $\delta < 1$.  This will imply that 
$\displaystyle 4 < x < 6 \implies 3 < (x-1) < 5$.
Therefore, $12 < 4(x-1) < 20.$ 
Therefore, the artificial constraint that $\delta < 1$ has established that the denominator in expression (1) above is greater than $(12)$.
As for the numerator in expression (1) above, 
the original constraint of $|x - 5| < \delta \implies$ that the absolute value of the numerator is less than $\delta$.
Therefore, under the assumption that $\delta < 1$, I have that
expression (1) above is less than 
$\displaystyle \frac{\delta}{12}$.
I want expression (1) above to be less than $\epsilon$. 
I could attempt to make a tight bound between $\delta$ and $\epsilon$.  I prefer to keep things simple.
Therefore, my candidate specification for $\delta$ is 
$\displaystyle \delta = \min[\epsilon, (1/2)].$
Now, I will verify that my candidate specification works.
I know that $|x - 5| < \delta = \epsilon.$
Since $\delta < 1$, I know that 
$4 < x < 6.$
This implies that $|4(x-1)| > 12.$
Therefore, expression (1) above is less than
$\displaystyle \frac{\delta}{12} < \delta = \epsilon.$
Therefore, the candidate specification of 
$\displaystyle \delta = \min[\epsilon, (1/2)]$ has been verified.
Therefore, the $\displaystyle \lim_{x \to 5} \frac{1}{x-1} = \frac{1}{4}.$
