Certain Limit is "Uniformly" Equal to $0$ I am reading through Conway's proof of Cauchy's integral formula and there are a couple claims he makes which I don't understand. In this post, I just want to investigate one claim. Here is the setup. Let $G$ be an open subset of $\Bbb{C}$ and $\gamma$ a closed, rectifiable curve in $G$. He claims that

$\lim_{z \to \infty} (w-z)^{-1} = 0$ uniformly for $w$ in $\{\gamma\}$...

where $\{\gamma\}$ denotes $\text{Im } \gamma$. By this, I am guessing he means that

for every $\epsilon > 0$, there exists $\delta > 0$ such that for every $w \in \text{Im } \gamma$, $|z| > \delta$ implies $\left|\frac{1}{w-z} \right| < \epsilon$.

I tried proving this, but I didn't have much luck. I was able to show that $\lim_{z \to \infty} \frac{1}{w-z} = 0$ for each $w \in \text{Im } \gamma$ (I guess that'd be pointwise convergence), but not uniform convergence. I could use some help.

$\operatorname{Im } \gamma$ is compact, and therefore contained in some disk $B_R(0)$. For $|z| > R$ and all $w \in \operatorname{Im } \gamma$ is, using the triangle inequality:
$$
 \left| \frac{1}{z-w} \right| = \frac{1}{|z-w|} \le \frac{1}{|z|-|w|}
\le \frac{1}{|z| - R} \, .
$$
Now, given $\epsilon > 0$, you can choose $\delta > R $ so large such that $\frac{1}{\delta - R} < \epsilon$.
