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Let $D:=\{z\in \mathbb{C}: \lvert z \rvert < 1\}$ and $B := \{z\in \mathbb{C}: \lvert z \rvert = 1\}$. Let $f:D\to \mathbb{C}$ be a uniformly continuous function. Then $\lim\limits_{z\to z_0, z\in D} f(z)$ exists for all $z_0\in B$.

I'm not sure how to approach this problem. Here's my guess: since points in $B$ are limit points of $D$, $\exists$ a sequence $\{z_n\}$ converging to $z_0 \in B$, for any $z_0\in B$. Now, since $f$ is continuous, $\lim\limits_{n\to \infty} f(z_n) = \lim\limits_{z\to z_0, z\in D} f(z)$ must exist. But if the proof were this straightforward, one could do away with simple continuity, but in this case uniform continuity is implied. So I think that I'm missing something.

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The problem seems to be the distinction between continuity and continuity at the boundary. Since the function is originally not defined at the boundary, you should not apply the definition of continuity at the boundary.

When you have a sequence converging to a point on the boundary, the sequence itself is Cauchy sequence. Uniform continuity gives you the image of the sequence being cauchy.

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    "Uniform continuity gives you the image of the sequence being cauchy." Can you please clarify this? – 2017-01-19
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    The tail of the sequence is bounded by an delta-ball, so the tail of the image is bounded by an epsilon ball – 2017-01-19
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    If $\{z_n\}$ is Cauchy and $f$ is continuous (not necessarily uniformly), then isn't it true that $f(z_n)$ is also Cauchy, at least in a complete metric space like $\mathbb{C}$ (since $f(z_n)$ converges)? – 2017-01-19
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    No, try $f(x)=1/x$ find a cauchy sequence $z_n \to 0$, what does $f(z_n)$ look like? – 2017-01-19
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    $1/x$ is continuous on its domain, but discontinuous at $0$. Couldn't we find a function that is not necessarily uniformly continuous, but is continuous on $\mathbb{C}$? Then it doesn't need to be uniformly continuous for $f(z_n)$ to be Cauchy, correct? – 2017-01-19
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    I thought your question says the domain is only $D$, the interior of the unit disc. – 2017-01-19
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    You are right. But if we find a function $g$ that is continuous on $D$, but not necessarily uniformly continuous on $D$, then isn't it true that $g(z_n)$ will be Cauchy as long as $z_n$ is Cauchy? – 2017-01-19
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    Like before, $f(x)=1/x$ is continuous on interval $(0,1)$ but not uniformly, continuity cannot extend to the boundary point $\{0\}$. – 2017-01-19
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    So if $f$ is uniformly continuous and $\{z_n\}$ is Cauchy then $f(z_n)$ is also Cauchy, which implies that $f(z_0)$ is continuous at the boundary, and hence the limit exists? Does this imply that the domain of $f$ can thus be extended to $D\cup B$? – 2017-01-19