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 $|f(x)|<M$ 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 $-L - \varepsilon <f(x) < L+ \varepsilon$ 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 $L-\varepsilon<f(x)<L+\varepsilon$ 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 $|f(x)|\leq |L|+\varepsilon$.

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  • $\begingroup$ The reverse inequality solution is brilliant! However can you elaborate more on why I don't need an arbitrary e? I'm asked to draw an illustrative diagram for the next part of this question, so I'm wondering if I should set e = 1 or stick with arbitrary e. What are the differences? $\endgroup$ – JaP Apr 14 '19 at 22:20
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Apr 14 '19 at 22:36

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