I have to prove the following using the epsilon delta method
$$ \lim_{x \to \,-1^{-}} \, \frac{5}{(x+1)^3} = - \infty $$
So, using the language of quantifiers, I think, I have to prove the following
$$ \forall N < 0 \,\, \exists \delta >0 \,\, \forall x \in D \,\Big[ -1 - \delta < x < -1 \Bigr] \longrightarrow \Biggl[ \frac{5}{(x+1)^3} < N \Biggr] $$
Now, playing with the consequent, I guessed the $\delta$ I needed here. So, following is my version of the proof.
Let $N < 0$ be arbitrary. Now choose
$$ \delta = \biggl( - \frac{5}{N} \biggr)^{1/3} $$
Since $N < 0$, we can see that $\delta > 0$. Now, let $x \in D$ be some arbitrary and then suppose that $ -1 - \delta < x < -1 $. After substituting for $\delta$, we have
$$ -1 - \biggl( - \frac{5}{N} \biggr)^{1/3} < x < -1 $$
From this, its clear that
$$ \biggl( \frac{5}{N} \biggr)^{1/3} < x + 1 < 0$$
$$\therefore \frac{5}{N} < (x+1)^3 < 0$$
$$ \Rightarrow \,-\frac{5}{N} > -(x+1)^3 > 0$$
$$ \Rightarrow \,-\frac{N}{5} < -\frac{1}{(x+1)^3} $$
$$ \Rightarrow \, \frac{5}{(x+1)^3} < N$$
Since $x \in D$ and $N < 0$ are arbitrary to begin with, this proves the limit using epsilon delta method.
Is this good enough ?