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Let $f:(0,1) \to \mathbb{R}$ be a given function and $\lim\limits_{x \to x_0} f(x) = L$. Then I believe the following definiton is equivalent to the definiton of the limit of a function $f(x)$:

for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $0 \lt |x-x_0| \lt \delta$, one has $0\lt |f(x) - L| \leq \epsilon$.

I think its equivalent because of $0 \lt |f(x)-L|$ i.e. the existence of the left side of the inequality doesn't matter for the definition of the limit of a function.

Am I right?

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This does not work because of the condition $0 < |f(x) - L|$.

Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 \not > 0$.

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  • $\begingroup$ that was my first guess, thanks for clarification . $\endgroup$ – ISuckAtMathPleaseHELPME Dec 16 '18 at 21:49
  • $\begingroup$ No problem. Feel free to accept if I definitively answered your question. $\endgroup$ – parsiad Dec 16 '18 at 21:50
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That's not equivalent since we can have $f(x)=L$ also for $x \neq x_0$ as for example for a constant function, therefore we should state

$$0\le |f(x) - L| \leq \epsilon$$

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