If $0 < |x-a| < \varepsilon$, and it's possible for $x=a$, then wouldn't the equation become $0 < 0 < \varepsilon$ (which is technically impossible)?
2026-04-11 20:12:25.1775938345
Definition of Limits: x=c
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Read carefully and you will see that the definition never says $0<|x-a|<\epsilon$.
What it says (and although this may seem like a small difference, it is vitally important) is that
In the case $x=a$, the definition is literally saying nothing at all - the other inequality may be true or it may be false.
The reason the definition is framed in this way is precisely so that the behaviour of the function when $x=a$ is irrelevant. A good way to understand why this is important is to consider the derivative of $f(x)$ when $x=a$, which is the limit of $$\frac{f(x)-f(a)}{x-a}$$ as $x\to a$. Clearly we must disallow the possibility $x=a$, as in this case the expression is $0/0$, which is meaningless.