# Proof Checking: Corollary epsilon delta definition of limit

I'm trying to solve this problem:

Suppose that $$\lim_{x \rightarrow c} f(x)$$ exists. Prove that there exists a constant $$M$$ and a $$\delta > 0$$ such that $$|f(x)| for $$0 < |x - c| < \delta$$.

My attempt is as follows:

From the $$(\varepsilon, \delta )$$-definition of the limit, we have that for every $$\varepsilon > 0$$, there exists a $$\delta$$ such if $$0 < |x - c | < \delta$$, then $$|f(x) - L| < \varepsilon$$. Then from the definition of the absolute value we have that $$-\varepsilon < f(x) - L < \varepsilon$$, and so $$L-\varepsilon < f(x) < L+\varepsilon$$. Furthermore since $$-L - \varepsilon < L- \varepsilon$$, we have that $$-L - \varepsilon or $$|f(x)| < L + \varepsilon = M$$ as required.

I'm specifically wondering if the part where I said $$-L - \varepsilon < L- \varepsilon$$ is valid, since $$L$$ could be negative?

As you say, you cannot do that step like that since $$L$$ could be negative. What you can do is, from $$L-\varepsilon you get $$|f(x)|\leq\max\{|L-\varepsilon|,|L+\varepsilon|\}\leq |L|+\varepsilon.$$
• Small criticism: you don't need an "arbitrary" $$\varepsilon$$. Just take $$\varepsilon=1$$ (say) to get $$|f(x)|\leq|L|+1$$.
• A more natural way to do the problem: use the (reverse) triangle inequality. So you have $$|f(x)|-|L|\leq|f(x)-L|<\varepsilon,$$ and thus $$|f(x)|\leq |L|+\varepsilon$$.
• You work with an "arbitrary" ε when your goal is to prove a universal statement ("for all ε"...). If you are looking for an estimate (an existential statement) havingan arbitrary $\varepsilon$ makes little sense: you are allowing it to be as big as you want, which kind of defeats the purpose of having an upper bound. – Martin Argerami Apr 14 '19 at 22:36