Example 2
\(\frac{1}{x}\) is continuous on \((0,1)\) but not uniformly continuous on \((0,1)\).
Proof \(1/x\) is continuous
Let \(\epsilon > 0\) be given. Let \(\delta = \min\{\frac{a^2\epsilon}{2}, \frac{a}{2}\}\). Now suppose that
$$
\begin{align}
|x - a| < \delta \quad \quad \quad (1)
\end{align}
$$
Then since \(|x - a| < \frac{a}{2}\), then by the reverse triangle inequality we have
$$
\begin{align}
|a| - |x| &\leq |x - a| < \frac{a}{2} \nonumber \\
a - \frac{a}{2} &\leq |x| \nonumber \\
\frac{a}{2} &\leq |x| \quad \quad \quad (2)
\end{align}
$$
Now observe that
$$
\begin{align*}
|f(x) - f(a)| &= \left| \frac{1}{x} - \frac{1}{a} \right| \\
&= \left| \frac{a - x}{xa} \right| \\
&= \frac{|a - x|}{|x|a} \\
&< \delta \frac{1}{|x|a} \quad \text{(By (1))} \\
&\leq \delta \frac{2}{a^2} \quad \text{(By (2))} \\
&\leq \epsilon \quad \text{(Since $\delta \leq \frac{a^2\epsilon}{2}$)}
\end{align*}
$$
Thus,
$$
\begin{align}
\lim_{x\to a} \frac1x = \frac1a.
\end{align}
$$
\(1/x\) is not uniformly continuous
To show that \(f\) is not uniformly continuous on \((0,1)\), we want to show that no choice of \(\delta\) could ever work for the entire interval without depending on \(a\). So we want to show that there exists some \(\epsilon_0 > 0\) such that for any \(\delta > 0\), there exists some \(x,y \in (0,1)\) with
$$
\begin{align}
|x - y| < \delta \text{ and } |f(x) - f(y)| \geq \epsilon_0
\end{align}
$$
Intuitively, \(\delta\) will fail near zero. Since near zero, we will see \(1/x\) blow up really high.
Proof
Pick \(\epsilon_0 = 1\). Now let \(\delta > 0\) be arbitrary. Then we have two cases: If \(0 < \delta \leq 1\), then let
$$
\begin{align}
x_{\delta} = \frac{\delta}{2} \quad \text{ and } \quad y_{\delta} = \frac{\delta}{4}
\end{align}
$$
Then
$$
\begin{align}
|x_{\delta} - y_{\delta}| = \left| \frac{\delta}{2} - \frac{\delta}{4} \right| = \frac{\delta}{4} < \delta
\end{align}
$$
Moreover, \(0 < y_{\delta} < x_{\delta} < \delta \leq 1\) so \(x_{\delta}, y_{\delta} \in (0,1)\). However
$$
\begin{align}
|f(x_{\delta}) - f(y_{\delta})| = \left|\frac{1}{x_{\delta}} - \frac{1}{y_{\delta}}\right| = \left| \frac{2}{\delta} - \frac{4}{\delta} \right| = \frac{2}{\delta} \geq 2 > \epsilon_0
\end{align}
$$
Otherwise \(\delta > 1\). But here just pick \(x = \frac{1}{2}\) and \(y = \frac{1}{4}\) to see that
$$
\begin{align}
|x - y| = \left| \frac{1}{2} - \frac{1}{4} \right| = \frac{1}{4} < 1 < \delta
\end{align}
$$
Moreover, \(0 < y < x < 1\) so \(x, y \in (0,1)\). However
$$
\begin{align}
|f(x) - f(y)| = \left|\frac{1}{x} - \frac{1}{y}\right| = \left| 2 - 4 \right| = 2 > \epsilon_0
\end{align}
$$
References