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)

The first is incorrect.
$$|x-2|<1\implies 1 <x <3\implies $$
$$\frac {1}{6}<\frac {1}{2x}<\frac {1}{2} $$
thus
$$|\frac {1}{x}-\frac {1}{2}|<\frac {|x-2|}{2} $$
if $|x-2|<1$ and $|x-2|<2\epsilon $ then $$|\frac {1}{x}-\frac {1}{2}|<\epsilon $$
we take $$\delta=\min (1,2\epsilon) $$
