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What is the importance of the difference between these two?

$$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x)([0 < \lvert x-a \rvert < \delta ] \Rightarrow [\lvert f(x)-L \rvert < \varepsilon ])$$ $$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x)([0 < \lvert x-a \rvert \leq \delta ] \Rightarrow [\lvert f(x)-L \rvert \leq \varepsilon ])$$

Thanks

Joseph
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1 Answers1

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Assume (1). Consider an arbitrary $\epsilon>0$. Then because $\epsilon>0$, there exists a $k> 0$ such that $$0 < \lvert x-a \rvert < k \Rightarrow \lvert f(x)-L \rvert < \varepsilon $$ In particular, when $\delta=k/2$, $|x-a| \le \delta$ implies $|x-a| < k$ implying $\lvert f(x)-L \rvert < \varepsilon$ implying $\lvert f(x)-L \rvert \le \varepsilon$. Thus (2) holds.

Conversely, assume (2). Consider an arbitrary $\epsilon>0$. Then because $\epsilon/2>0$, there exists a $k> 0$ such that $$0 < \lvert x-a \rvert \le k \Rightarrow \lvert f(x)-L \rvert \le \varepsilon/2 $$ In particular, when $\delta=k$, $|x-a| < \delta$ implies $|x-a| \le \delta$ implying $\lvert f(x)-L \rvert \le \varepsilon/2$ implying $\lvert f(x)-L \rvert < \varepsilon$. Thus (1) holds.

user156213
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  • Thank you, this clears it up for me. That was my general intuition but now I can see how it works out. – Joseph Oct 12 '18 at 06:26