Epsilon-Delta of a cubic function

Question

I have the cubic function $$ \lim_{x \to -2}(x^3+4x^2+4x-1)=-1 $$ which I manage to break down into $$x(x+2)^2 < \epsilon.$$

How do I proceed with this? I try to define $x$ by setting $\delta < 1$ but get negative outputs which I frankly do not know what to do with.

Answer 1

So, we want from $|x-(-2)|=|x+2|<\delta$ conclude $|x(x+2)^2|<\varepsilon$ and problem seems to estimate $|x|$, as for $|x+2|$ estimation is clear.

Obviously $|x+2|<\delta \Leftrightarrow 2-\delta<x<2+\delta$, so taking $\delta<1$ gives $1<|x| =x < 3$. Can you finish from here?

Answer 2

$$ \lim_{x \rightarrow -2} (x^3 +4x^2 +4x -1)=-1\\ \iff |(x^3 +4x^2 +4x -1)-(-1)| = |x^3 +4x^2 +4x| = |x(x^2 +4x +4)| = |x(x+2)^2| \leq |x||x+2|^2 < \varepsilon \\ \Longleftarrow (|x| < \delta_0\ \wedge\ |x+2|^2 < (\varepsilon/\delta_0) = \delta_1^2)\\ $$ where $\delta_0$ is any positive number, no matter how large. The last conjunct can be written $$ |x+2| < \delta_1 = \sqrt{\varepsilon/\delta_0} $$ Taking $\delta_0$ to be the small one implies that the function has the same limit at $x=0$.