Meaning of uniform in "$f$ tends to $L$ uniformly as $z$ tends to $z_0$ through a cone"

Question

Let $f$ be an analytic function on the unit disk, and let $z_0$ be a point on the boundary. I would like to know the precise meaning of the wording by Gamelin

$f(z)$ tends uniformly to $L$ as z tends to $1$ through any cone with vertex at $1$ contained in the unit disk.

I understand what it means for a function to be uniformly continuous on a domain ($\delta$ is independent of $z$ given $\epsilon$), and what it means for a sequence of functions to converge uniformly on a domain ($n(\epsilon)$ is independent of $z$ given $\epsilon$). Alas uniform convergence of a sequence of functions speaks of the convergence at each point of the domain, and here it is referred to the behaviour when tending to a point $z_0$.

Also, is it equivalent to just saying that the analytic function tends to $L$ as z tends to $1$ through any cone, Which is just convergence through every sequence of points in the cone tending to $z_0$?

Or is uniformity something nontrivial that should be proved on top of proving convergence?

ADDED

Gamelin also asks to show that if $f$ is a bounded analytic function on a horizontal strip $-1<\text {Im}(z)<1$ and $\lim f(x)=L$, then $\lim _{x \to \infty}f(x+iy) = L$ uniformly for $-1+\epsilon<y<1-\epsilon$.

Does he mean by "uniform" that there is $M(\epsilon)$ independent of $y$ for which $|f(x+iy)-L|<\epsilon$ for all $x>M$?

Answer 1

The convergence is just the usual convergence of a function at a region, and the "uniform" adjective means nothing in Gamelin's chapter XI.5, exercises $4$ and $5$.

In the other mentioned exercise Gamelin does not mean the phrase that I highlighted at the end of the post, but rather a convergence in the region $\{x \in \Bbb R, y \in (1-\varepsilon, -1+\varepsilon)\}$.

Answer 2

The auther assumes (or hopes) a reader of his book solves exercises $3,4$ and $5$ using the second theorem at page 308 which states that $\mathcal{F}$, a family of analytic functions uniformly bounded on each compact subset of $D$, has a subsequence that converges uniformly on each compact subset of $D$. So the reader succeeds to solve them, uniform convergence is automatically fulfilled.

Suppose that you could prove (by another way different above) that $f(z)$ tends to $L$ as $z$ tends to $1$ through a cone $\Delta$ with vertex at $1$ contained in the unit disk, without mentioning about uniformity of convergence.
Then define $f(1)=L.$ So $f(z)$ will be continuous on $\overline{\Delta}$, the closure of $\Delta$.
Since $\overline{\Delta}$ is compact, $f(z)$ is uniformly continuous on it. Therfore $f(z) \to L,\, (z\to 1, z\in \Delta)$ is uniform, even if you did no reference.

We have the same aspect in case of a horizontal strip, when a horizontal strip is mapped to the unit disk.