Proving $\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $ using $\varepsilon$-$\delta$ 
Prove that $$\lim_{x\to 1}\sqrt{2x} = \sqrt 2 $$ using $\varepsilon$-$\delta$. 

What I've tried is 
$$\left|\frac{(\sqrt{2x}-{\sqrt 2})(\sqrt {2x}+{\sqrt 2})}{\sqrt {2x}-{\sqrt 2}}\right| 
=\left|\frac{2x-2}{\sqrt {2x}-{\sqrt 2}}\right|$$ and because $|x-1|< \delta$ implies that $x> 0$ so $\sqrt {2x}$ is defined.
$$\frac{|2||x-1|}{\sqrt 2}<\varepsilon$$ then we can choose
$$\delta= \frac{\varepsilon \sqrt 2 }{2}.$$
 A: You're aiming to make $|\sqrt{2x} - \sqrt{2}|$ very small. Rationalising the numerator is a good idea, but you seemed to rationalise $|\sqrt{2x} + \sqrt{2}|$ instead. Rather, we have
$$|\sqrt{2x} - \sqrt{2}| = \left|\frac{(\sqrt{2x} - \sqrt{2})(\sqrt{2x} + \sqrt{2})}{\sqrt{2x} + \sqrt{2}}\right| = \frac{|2x - 2|}{\sqrt{2x} + \sqrt{2}}.$$
Now, note that, for all $x \ge 0$,
$$\sqrt{2x} \ge 0 \implies \sqrt{2x} + \sqrt{2} \ge \sqrt{2} \implies \frac{1}{\sqrt{2x} + \sqrt{2}} \le \frac{1}{\sqrt{2}}.$$
Therefore,
$$|\sqrt{2x} - \sqrt{2}| = \frac{1}{\sqrt{2x} + \sqrt{2}}|2x - 2| \le \frac{1}{\sqrt{2}}|2x - 2| = \sqrt{2}|x - 1|.$$
So, if we force $|x - 1| < \frac{\varepsilon}{\sqrt{2}}$, then we have
$$|\sqrt{2x} - \sqrt{2}| \le \sqrt{2}|x - 1| < \varepsilon,$$
as necessary. This suggests choosing $\delta = \frac{\varepsilon}{\sqrt{2}}$, as you did in your working.
However, it is worth noting that such a choice of $\delta$ may not ensure $x$ is positive. In particular, if we look at $\varepsilon = 4\sqrt{2}$, then the corresponding value of $\delta$ would be $4$. This is too big as the number $-1$ is within distance $4$ of $1$, but is not in our domain. So, we should probably limit $\delta \le 1$, in order for $(1 - \delta, 1 + \delta)$ to consist only of positive numbers. That is, choose
$$\delta = \min\left\{1, \frac{\varepsilon}{\sqrt{2}}\right\}.$$
