I have to prove the following using epsilon delta: $$\lim\limits_{x \to -2} x^3+4x^2+4x-1 = -1$$
if $|x+2| < \delta$ then $|f(x) - f(a)| < \varepsilon$
Factorizing f(x): $|(x+2)(x^2+2x)| < \varepsilon$
Divide both sides by $|x^2+2x|$ and get: $|x+2| < \frac{\varepsilon}{|x^2+2x|}$.
I have $-\delta < x+2 < \delta$ from $|x+2| < \delta$
I then set $\delta:=1$ and subtract $2$ from all sides: $-3 < x < -1$
I do this to create an upper bound for $x$ and therefore an upper bound for $|x^2+2x|$
How do I know which side of the inequality (-3 or -1) to substitute for $x$ in $\frac{\varepsilon}{|x^2+2x|}$?
Alternatively, if thats the wrong approach, what's the right approach for deciding $\delta$?
For any $x$ such that $|x+2|<1$ we have that $|x|<3$, and so, for all $\varepsilon>0$, if $x$ satisfies that $$|x+2|<\min\Big\{1,\sqrt \frac{\varepsilon}{3} \Big\}$$ then $\displaystyle |(x^3+4x^2+4x-1)-(-1)| = |x(x+2)^2| = |x||x+2|^2 < 3 \Big( \sqrt \frac{\varepsilon}{3} \Big)^2 = \varepsilon.$