Prove, using the epsilon-delta definiton, the following limit exists：
$$\lim_{x \to 0}{\sqrt{4 - x}} = 2$$
2026-03-31 05:48:58.1774936138
Use delta and epsilon to prove the limit
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Hint: $|\sqrt{4-x}-2|=|\frac{(4-x)-4}{\sqrt{4-x}+2}|\leq\frac{|x|}{2}$. If $|x|<2\varepsilon$, then we get the convergence as $|\sqrt{4-x}-2|<\varepsilon$.