Show that if $ \lim_{x \to a} f(x) = -\infty$ then $\lim_{x \to a} (f(x))^2 = \infty$ 
Let $a \in \mathbb{R}$. Use the $\epsilon-\delta$ definition of the limit to show that if $\displaystyle \lim_{x \to a}f(x) = -\infty$ then $\displaystyle \lim_{x \to a} (f(x))^2 = \infty$.

We are given that $$\forall M < 0, \exists \delta \quad 0 <|x-a| < \delta \quad \implies \quad f(x) < M$$ and need to show $$\forall M > 0, \exists \delta \quad 0<|x-a|<\delta \quad \implies \quad (f(x))^2 > M$$. How can I get from the first to the second? 

The two $M$s can be distinct provided they can be arbitrarily large. Here is how I would write it.
$\forall\,M > 0, \exists\,\delta > 0, |x - a| < \delta \Rightarrow f(x) < -M \iff -f(x) > M \Rightarrow (f(x))^{2} > M^{2}$.
