# equivalent definition of limit of a function?

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?

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 $$x$$, $$|f(x) - L| = |0 - 0| = 0 \not > 0$$.
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
$$0\le |f(x) - L| \leq \epsilon$$