Continuity and uniform continuity of $f(z)=\frac{1}{z}$? Real Analysis experts ,
I have learnt that $f(z)=\frac{1}{z}$ is not uniformly continuous on $0<z\le1$.
My question is:
Is $f(z)=\frac{1}{z}$ uniformly continuous on $0<z<1?$
Thanks for your guidance.
 A: Hint: $\left|f\left(\frac 1 {2^n}\right)-f\left(\frac 1 {2^{n+1}}\right)\right|=2^n$.
A: A useful theorem, A continuous function on $(a,b)$ , where $a,b \in \mathbb{R} $  is uniformly continuous on $(a,b)$ iff $\lim_{x \to a+ } f(x) $ and $\lim_{x \to b- } f(x) $ exists finitely.
In your problem, $\lim_{z\to 0+ } \frac{1}{z} \to +\infty $
Edit: similar case also for complex valued function, if  a complex valued function $f$ is continuous on the region $a \lt |z| \lt b $ , then $f$ is uniformly continuous on $a \lt |z| \lt b $  iff $\lim_{z \to z_{0} } f(z) $ exist finitely, where $z_{0} $ runs over all limit points of the region $\{z : a \lt |z| \lt b \} $ ,
You can check that $\lim_{z \to 0 } \frac{1}{z} $ doesn't exist if we approach $0$ along real line,
A: For a more geometrical approach, see the intuitive example provided by Wikipedia:

In the above GIF, the function $f(x)=\frac{1}{x}$ is continuous but not uniformly continuous. This is because the graph of $f(x)$ escapes the top and/or bottom of the $\text{height}\times \text{width} = 2\varepsilon \times 2\delta$ window irregardless of how small $\delta$ is. In contrast, the function $g(x)=\sqrt{x}$ is uniformly continuous as it never escapes the box.

The function $f(z)=1/z$ is continuous everywhere it's defined, that is, everywhere except $z=0$.
But it's not uniformly continuous in a punctured neighborhood of $0$. In fact, it's the standard example of a function that's continuous on the open interval $(0,1)$
but not uniformly continuous there.
By contradiction, suppose that $f(z)$ is uniformly continuous in the region $0<|z|<1.$ Then, for all $\varepsilon >0$, there exists a $0<\delta<1$ such that $|z_1-z_2 |<\delta \implies \left|f(z_1)-f(z_2)\right|<\varepsilon$ where $z_1,z_2$ are arbitrary elements in the domain of $f$.
Let $z_1=\delta$ and $z_2=\delta/(1+\varepsilon)$. Then
$$|z_1 - z_2| = \frac{\delta\varepsilon}{1 + \epsilon} < \delta,$$
but
$$\left|f(z_1) - f(z_2)\right| = \left|\frac{1}{\delta}-\frac{1 + \epsilon}{\delta} \right| = 
\frac{\epsilon}{\delta} > \epsilon,$$
which is a direct contradiction to the definition of uniform continuity. Therefore, $f(z)$ is not uniformly continuous on the unit punctured disk.
