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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?

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  • $\begingroup$ $\forall M>0 \;\exists \delta >0\; (0<|x-a|<\delta \implies f(x)<-\sqrt M).$ $\endgroup$ – DanielWainfleet Mar 7 '16 at 19:21
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The two $M$s can be distinct provided they can be arbitrarily large. Here is how I would write it.

$\forall\,M > 0, \exists\,\delta > 0, |x - a| < \delta \Rightarrow f(x) < -M \iff -f(x) > M \Rightarrow (f(x))^{2} > M^{2}$.

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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)

$$ f(x)^2 = |f(x)|^2 > m^2 = M. $$

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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), $f(x)^2> Mf(x)$.

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