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Just need to make sure the proof makes sense, I suppose

We want to prove that a function has a limit at point $x_0$, then that limit is unique.

Let f: $E \rightarrow Y$, an application of E to Y, with X and Y metric spaces, and $E \subset X$. Let $x_o$ be a cluster point of E, and L and G two limits of f at $x_o$.

By definition we have: $$ \forall(u_n)_{n\in \mathbb N} \in E, s.t. \lim_{n\to \infty } u_n = x_o $$

$$ \lim_{n\to \infty}f(u_n) = L $$

And

$$ \forall(u_n)_{n\in \mathbb N} \in X, s.t. \lim_{n\to \infty } u_n = x_o $$

$$ \lim_{n\to \infty}f(u_n) = G $$

Immediately we conclude that $L =G$

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    The proof doesn't make sense to me because I don't know what it is proving. It's common to first write the statement you want to prove, otherwise readers get confused. I also have no idea what "continuous open intervals" are. – 2017-02-22
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    @5xum Edited for clarity, thank you :) – 2017-02-22
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    Still don't know what a "continuous open interval" is. "Continuous" is a property of a *function*, not a *set*. – 2017-02-22
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    @5xum Fixed that too – 2017-02-22

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The definition for limits of function you're using differ quite a bit from the standard definition. The definition you're using is not a very practical definition in general. In general when you're going to prove something about limits you would want the delta-epsilon definition and with your definition for limit you would basically prove that the delta-epsilon-limit is implied.

What you should do if you were using the standard definition is prove it from that. Suppose that $f(x)\to L$ and $f(x)\to M$ (this notation is fine as it doesn't rely on limits being unique). Assume also that $L\ne M$ which means that $|L-M|>0$. Now we see that the definition does not work since if we pick $\epsilon < |L-M|/3$ we can't find $\delta$ to ensure that both $|f(x)-L|<\epsilon$ and $|f(x)-M|<\epsilon$. In fact $3\epsilon < |L-M| = |f(x)-M+L-f(x)| \le |f(x)-M| + |f(x)-L|$ which contradicts the premisses and therefore we can conclude that $L\ne M$ is false.

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    @skying Isn't the definition of limits using sequences just as legitimate as the delta-epsilon one? They are equivalent... Should I just not want to prove the uniqueness of a limit using that definition? Also, the limit of a sequence is well defined in this context, which is why we can use it to construct the limit of a function. – 2017-02-22
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    @Quantaliinuxite That's very much an illusion. Even if you define the limit using sequences you just moves the very definition to that of limits of sequences which in turn uses the epsilon definition. I don't think it's very common to use such definition of limits. Anyway if done that way the uniqueness would have to be proven for sequences. – 2017-02-22
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    @Quantaliinuxite What's "a definition of limits using sequences"? Can you provide a strict mathematical wording of this definition? – 2017-02-23
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    @5xum $f(x)\to L$ as $x\to a$ if for all sequences $x_n$ such that $\lim x_n = a$ we have $lim f(x_n) = L$. I don't think it's a practical definition, but as far as I can see it's equivalent to the normal definition. It's not practical since I imagine that every proof would rely on a theorem that says the normal definition would be implied (and use that instead). – 2017-02-23
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    @skyking But that's not a definition of **limits**. It's a definition of **continuity**. And yes, it is equivalent to the definition of continuity using $\epsilon-\delta$. But the question you are solving is not about continuity, so this definition is useless. – 2017-02-23
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    @5xum No, I don't think so. Keep in mind that this definition is **not** to be interpreted as given after the standard definition of limit of function, but **instead** of it. That is before this definition the statement $f(x)\to L$ as $x\to a$ didn't have a meaning, but after it it has the defined meaning. The definition of continuity is that the limit coincides with the value of the function, but that's not claimed here. Note that depending on the version of limit you want you could in the above definition require that eventually $x_n\ne a$. – 2017-02-23
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    @skyking OK, you are half right, but still you have, on the right side of your definition, the term $\lim$, and this term (for *sequences*) is defined through $\epsilon-\delta$. – 2017-02-23
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    @5xum Yes, of course, this definition relies on the definition for limits for sequences being in place first. It's also assumed that the one has already concluded that limits for sequences are unique. – 2017-02-23