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? 

Since the limit of f is negative infinity, the first part is true for negative M- in fact, for M< -1. In that case if f(x)< M both M and both f(x) and M are negative so that, multiplying both sides of f(x)< M by M, $Mf(x)> M^2$.  Then, multiplying both sides of f(x)< M by f(x), $f(x)^2> Mf(x)$.
