Consider$$f(x) = \frac{2\sin(x) + \sqrt x}{x}$$
When $x$ approaches zero, $\frac{2sinx}{x}$ goes to $2$, but $\frac{\sqrt x}{x}$ goes to infinity, whose graph indicates that its derivative when $x$ approaches zero goes to negative infinity. It suggests that no single $\delta$ should be able to respond to all $\epsilon$ s' challenges, thus $f(x)$ is not uniformly continuous.
To prove rigorously, my intention was to choose two series $x_n$ and $y_n$ so that |$x_n - y_n$|$\to 0$ and |$f(x_n) - f(y_n)$|$\ge \epsilon_0$, where $\epsilon_0 $ is chosen and $\gt 0$. But it turned out that it could be very hard to prove $|f(x_n)-f(y_n)|\ge$ any fixed number.
However, I have no idea how to apply another strategy, i.e. choose $\epsilon = \epsilon_0$, given any $\delta \gt0$, choose particular $x$ and $y$ s.t. $|x-y|\lt\delta$ and $|f(x)-f(y)|\ge\epsilon_0$. I tried to set $x = min\{1,\delta\}$ and $y = \frac x2$, it doesn't work either.
Could you guys help me out of this? I am fresh to analysis and this question has stung me for a while.
It is not uniformly continuous on $(0,a)$ for any $a >0$. If it is uniformly continuous then it would be be bounded as $x \to 0+$. But $\frac {\sin x} x \to 1$ as $ x\to 0+$ and $\frac 1 {\sqrt x} \to \infty$ as $x \to 0+$.
If you don't want to use the boundeness criterion you can argue as follows: Supppose there exists $\delta >0$ such that $|f(x)-f(y)| <1$ whenever $|x-y| <\delta $. Then there exists $n_0$ such that $|f(\frac 1 n)-f(\frac 1 m)| <1$ whenveer $n,m \geq n_0$. Put $m=n_0$ and take limit as $ n \to \infty$ to get a contradiction.