<s>[INST] I got a negative delta when computing deltas algebraically I am trying to find $\delta$ of $f(x)=4-x^2$ for $c=-1, L=3, \epsilon=\frac{1}{4}$. Here is what I did:
$$|f(x)-L|<\epsilon$$
$$|4-x^2-3|<\frac{1}{4}$$
$$-\frac{1}{4}<1-x^2<\frac{1}{4}$$
$$-\frac{5}{4}<-x^2<-\frac{3}{4}$$
$$\frac{5}{4}>x^2>\frac{3}{4}$$
$$\frac{\sqrt{5}}{2}>x>\frac{\sqrt{3}}{2}$$
$$\frac{\sqrt{3}}{2}<x<\frac{\sqrt{5}}{2}$$
Then the second step:
$$|x-c|<\delta$$
$$|x+1|<\delta$$
$$-\delta<x+1<\delta$$
$$-\delta-1<x<\delta-1$$
Left substitution:
$$-\delta-1=\frac{\sqrt{3}}{2}$$
$$\delta+1=-\frac{\sqrt{3}}{2}$$
$$\delta=-1-\frac{\sqrt{3}}{2}$$
Right substitution:
$$\delta-1=\frac{\sqrt{5}}{2}$$
$$\delta=\frac{\sqrt{5}}{2}+1$$
As you can see, I got a negative delta when doing the left substitution of the inequality ($\delta=-1-\frac{\sqrt{3}}{2}$). I heard it said somewhere that if you get a negative delta, you have an error in your algebra. Is this true? If so, what did I do wrong?
 [/INST] Hint:
Your passage : $\dfrac{5}{4}<x^2<\dfrac{3}{4} \rightarrow \dfrac{\sqrt{5}}{2}<x<\dfrac{\sqrt{3}}{2}$ is wrong. You have two second degree inequalities:
$$
{5}{4}<x^2 \qquad \land \qquad x^2<\dfrac{3}{4}
$$
and you have to solve separetely.
The correct solution is:
$$
-\dfrac{\sqrt{5}}{2}<x<-\dfrac{\sqrt{3}}{2} \quad \lor \quad \dfrac{\sqrt{3}}{2}<x<\dfrac{\sqrt{5}}{2}
$$
</s>