epsilon-delta proof of $\lim_{x \to 4} \sqrt{x} = 2$
Answer to your where did $\geq2 > 1$ comes from:
It comes from the fact $\sqrt{x} \geq 0$, so $|\sqrt{x} + 2| \geq |0 + 2| = 2$ and of course $2 > 1$.
epsilon-delta proof of $\lim_{x \to 4} \sqrt{x} = 2$
Answer to your where did $\geq2 > 1$ comes from:
It comes from the fact $\sqrt{x} \geq 0$, so $|\sqrt{x} + 2| \geq |0 + 2| = 2$ and of course $2 > 1$.