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)

For the second method, it is ok since the rational function is continuous on it definition area.    
