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so I am preparing for my final and I needed some help verifying my definitions and their justifications. So any help/feedback would be appreciated...

Let $f:(0,1) \to \mathbb{R}$ be a given function, and the limit $\lim\limits_{x \to x_0} f(x) = L$. Then out of the following defnition pick the right definition, and/or equuivalent definition. If the definition is not equivalent provide explanation:

a.) 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 $|f(x) - L|\lt \epsilon$.

this definition is true.

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

this definition is false. (I don't quite understand how tho)

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

this definition is false because for any ϵ>0 there exists some δ>0 that is small enough. It can't be any delta.

d.) 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 $0 \lt |f(x) - L|\leq \epsilon$.

this definition is false. suppose $f(x) = 0$ then with $L = 0$ the limit does not exist anywhere for every$x$ as $|f(x) - L| = 0$ $0$ is not less then $0$.

these are my attempts on the identification. Could you please help me confirm these answers?

Thank you.

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That one

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

is not correct, indeed by the condition

$$|x-x_0| \leq \delta$$

we could be allowed to take $x=x_0$ but in the definition we are assuming a deleted neighborhood of $x_0$ that is $x\neq x_0$.

The correct version is

b.2) for any $\epsilon \gt 0$, there exists $\delta \gt 0$ such that for all $x \in (0,1)$ and $\color{red}{0<}|x-x_0| \leq \delta$, one has $|f(x) - L|\leq \epsilon$

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    $\begingroup$ but, since we are not given $0 \lt$ part of the inequality, then it makes the definition invalid, right? $\endgroup$ – ISuckAtMathPleaseHELPME Dec 16 '18 at 23:06
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    $\begingroup$ oh okay, you edited after i commented but i got it thanks :). does c and d make sense tho? $\endgroup$ – ISuckAtMathPleaseHELPME Dec 16 '18 at 23:07
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    $\begingroup$ @ISuckAtMathPleaseHELPME Yes exactly, the first version was a little bit confusing! The key is that we are considering a deleted neighborhood of $x_0$. $\endgroup$ – gimusi Dec 16 '18 at 23:08
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Note that for French b) is the right definition

$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$

For other country (UK perhaps) a) is the right definition

$\forall \varepsilon>0,\;\exists \delta>0,\; \forall x\in (0,1),\quad \big[0<|x-x_0|<\delta \implies |f(x)-L|<\varepsilon\big]$

and $b\implies a$

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