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? 

It's enough to show that $\lim_{x\to a}f(x)^2$ exists (in $\overline{\mathbb{R}} = \mathbb{R}\cup\{\pm\infty\}$) and exceeds any positive $M$. So, fix $M > 0$ and let $m = \sqrt{M}$. By hypothesis, there exists $\delta > 0$ such that $0 < |x-a| < \delta$ implies $f(x) < -m$. Hence $0 < |x-a| < \delta$ implies $|f(x)| > |-m| = m$ and (hence)
$$
    f(x)^2 = |f(x)|^2 > m^2 = M.
$$
