Is the following proof of a limit valid? Although the following limit is extremely easy to prove but I want to know whether the following method is correct or not.
To prove 
$$\lim_{x \to 5}  (x-1)^{1/2}=2$$
So I have to prove the following implication $$\forall \varepsilon >0 $$ $$\exists \delta >0 : 0<\left | x-5 \right |<\delta \Rightarrow |(x-1)^{1/2}-2|<\varepsilon$$
Thus I have to prove the above implication for all epsilon>0.
Now consider the following chain of arguments for all epsilon:-
$$|(x-1)^{1/2}-2|\geq \varepsilon\\
|(x-1)^{1/2}-2||(x-1)^{1/2}+2|\geq \varepsilon|(x-1)^{1/2}+2|\\
|x-5|\geq \varepsilon|(x-1)^{1/2}+2|\\
\nexists \delta >0 : 0<\left | x-5 \right |<\delta$$
The last implication follows from the one before that because $$|x-5|\geq \varepsilon|(x-1)^{1/2}+2|$$ for all epsilon means that |x-5| is not bounded above.
Thus this chain of arguments has made the following implication vacuous and hence true
 $$\exists \delta >0 : 0<\left | x-5 \right |<\delta \Rightarrow |(x-1)^{1/2}-2|<\varepsilon$$
Which was required to be proven. I asked one of my friends and he said that I cant consider my chain of arguments for all epsilon, but since epsilon is not a free variable and hence is a bounded variable thus the truth value of my first statement in the chain is either 0 or 1 and hence the reasoning is not flawed. 
Or am I applying logic absurdly and I am confused.
Kindly help.
Thank you for reading. 
 A: You have shown $(\forall\epsilon>0$, $|(x-1)^{1/2} - 2| \geq \epsilon) \implies |x-5|$ is not bounded above $\implies \not\exists\delta>0: 0 < |x-5| < \delta$
But you need to show $\forall\epsilon>0$, $(\exists\delta>0$, such that $(0 < |x-5| < \delta \implies |(x-1)^{1/2} - 2| < \epsilon))$
Notice the placement of the parentheses. Do you understand why what you showed is not logically equivalent?
A: For all those kind of continuity questions, you have to FIX $\epsilon >0$ and prove the existence of $\delta$ (that depends on the fixed $\epsilon$) such that for $0 < \vert x- x_0\vert < \delta$ you have $\vert f(x) - f(x_0) \vert < \epsilon$.
The good point that you noticed is that:
$$\vert \sqrt{x-1} - 2 \vert = \frac{(\sqrt{x-1} - 2 )(\sqrt{x-1} + 2)}{\sqrt{x-1} + 2}.$$
That helps a lot because now you can state that
$$\vert \sqrt{x-1} - 2 \vert = \left\vert \frac{x-5}{\sqrt{x-1} + 2}\right\vert.$$
As $\sqrt{x-1} \ge0$, you get
$$\vert \sqrt{x-1} - 2 \vert \le \frac{\vert x- 5 \vert}{2}$$ and you're done if you notice that when choosing $\delta = 2 \epsilon$, you get that
$$\vert \sqrt{x-1} - 2 \vert \le \frac{\vert x- 5 \vert}{2} < \epsilon$$ when
$\vert x-5\vert < \delta = 2 \epsilon$.
