Limit on the unit disk

Let $A \subset \mathbb{R}^2$ be the open unit disk $D_1(0,0)$ with the point $\mathbf{x}_0 = (1,0)$ added, and let $f: A \to \mathbb{R}, \, \mathbf{x} \mapsto f(\mathbf{x})$ be the constant function $f(\mathbf{x})=1$.

How do I prove that $$\lim_{\mathbf{x}\to\mathbf{x}_0} f(\mathbf{x})=1$$ ?

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Well, such limit is defined only over the sequences which lie inside $A$. Also, for any $\{x_n\}\subset A$ we have $\lim_n f(x_n) = 1$ since $f\equiv 1$ is a constant function. We only have to check that $x_0\in \bar A$ for the limit to be well-defined.

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Given any $\epsilon > 0$, pick $\delta > 0$ arbitrarily. Then for all $x \in A$ with $|x - x_0| < \delta$, we have $f(x) = 1$, so $|f(x) - 1| = 0$. This means $\lim_{x\to x_0} f(x) = 1$.

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...and, what...?...If you'd like to answer, perhaps you can expand a bit. As is, this is not much of an answer. If you don't want to expand on this, please make this a comment below the post, and delete the answer. – amWhy Jan 28 '13 at 15:11
Ok. I will expand it. – Tunococ Jan 29 '13 at 0:29