<s>[INST] Proof verification: $\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$ Is either of the following methods correct?
Prove
$$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
First Method
Preliminary Analysis:
We know that $a = 2$, $L= \frac{1}{2}$, and $f(x)= \frac{1}{x}$. By the precise definition of limit we have the following:
$$ 
0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\\
0<|x-2|<\delta \implies \left|\frac{1}{x}-\frac{1}{2}\right|<\varepsilon \implies \left|\frac{2}{2x}-\frac{x}{2x}\right| <\varepsilon \implies \left|\frac{2-x}{2x}\right| <\varepsilon \\
\implies \left|\frac{-(-2+x)}{2x}\right| <\varepsilon \implies \left|\frac{x-2}{2x}\right| <\varepsilon \implies \frac{\left|x-2\right|}{\left|2x\right|} <\varepsilon \implies \left|x-2\right|<\varepsilon \left|2x\right|
$$
let $\delta = \varepsilon\left|2x\right|$ but we need to simplify it further because delta should be in terms of $\varepsilon$ only. 
Assume $|x-a| < 1$
$$ |x-2| < 1 \implies -1 <x -2<1 \implies -1+2<x<1+2 \implies 1<x<3$$
Then we have to simplify $|2x|$ as well, which ends up being 
$$ 2<2x<6 \implies -6<2x<6 \implies |2x| <6$$
Now consider the inequality we discovered, specifically, $\left|x-2\right|<\varepsilon \left|2x\right|$ this inequality is also valid for $\left|x-2\right|<\varepsilon\cdot6$. 
Let $\delta = \min{\{1, \varepsilon\cdot6}\}$
Proof:
Given $\varepsilon > 0$ let $\delta = \min{\{1, \varepsilon\cdot6}\}$ if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$ 
We also have  $|x - 2| < \varepsilon \cdot6$.
$$ 
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$
By the precise definition of limit $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Second Method
Proof by Definition/Property: 
By the direct substitution property if $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then 
$$\lim_{x \to a} f(x) = f(a)$$
Then by the direct substituon property of limit: $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Are either of the above methods correct? (I am putting the question here as well in case someone misses it)
 [/INST] proof-verification:

(1)Given $\varepsilon > 0$, let $\delta = \min{\{1, 6\varepsilon}\}$. 
(2)if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$ 
(3)We also have  $|x - 2| < \varepsilon \cdot6$.
$$ 
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$

The writing of these line 2 and 3 are terrible. One should not write those confusing big "implication" arrows for doing simple algebra. Moreover, (3) is incorrect: $|x-2|<6\varepsilon$ and $|2x|<6$ do not imply that
$$
\frac{|x-2|}{|2x|}<\frac{6\varepsilon}{6}.
$$ 

By the precise definition of limit, $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$


You want to get the estimate like
$$
\frac{|x-2|}{|2x|}<\epsilon.
$$
The intuition is as follows. On the one hand, $|x-2|$ can be as small as possible if $x$ is close to $2$. One the other hand, when $x$ is close to $2$, the quantity $\frac{1}{|2x|}$ is bounded by some positive real number. Thus together, one can get $\frac{|x-2|}{|2x|}$ as small as one wants when $x$ is close to $2$. 
Now we turn the intuition to a rigorous proof. Given $\epsilon>0$, let $\delta = \min\{1,\epsilon\}$. If $0<|x-2|<\delta$, then $1<x<3$, which implies that
$$
\frac{1}{|2x|}<1 \tag{1}
$$
On the other hand, by the definition of $\delta$, we also have
$$
|x-2|<\epsilon.\tag{2}
$$
Combining (1) and (2), we have the desired inequality
$$
\frac{|x-2|}{|2x|}<\epsilon. 
$$
</s>