Epsilon-delta limit proof, but delta turns out to be negative

Question

Have to prove that $$\lim_{x \to 0} \sqrt{4-x} = 2$$ Using epsilon-delta definition. So have to show that $\forall\varepsilon>0, \exists\delta>0$ such that $0<|x|<\delta \Rightarrow |\sqrt{4-x}-2|<\varepsilon$. When you work it out you get that $\delta = -\varepsilon^2+4\varepsilon$. What I don't understand is that, if you plug in $\varepsilon=5$ you get $\delta = -5$. But 1) what does it even mean to have negative $\delta$ and 2) you should be able to plug in any nonnegative $\varepsilon$ and get a nonnegative $\delta$ by definition, so why does it not happen in this case?

Answer 1

Since you did not post how you found expression for $\delta$, here is what I think:$|\sqrt{4-x}-2|=|\frac{-x}{\sqrt{4-x}+2}|<|\frac{\delta}{\sqrt{4-x}+2}|<|\frac{\delta}{2}|=\frac{\delta}{2}<\epsilon \implies \delta = 2\epsilon$