Is $\sqrt{x}(\sin{1\over x}+x)$ uniformly continuous on $(0,\infty)$?

Question

Is $\sqrt{x}(\sin{\dfrac{1}{x}}+x)$ uniformly continuous on $(0,\infty)$?

At first, I tried to take the derivative and show it is bounded, but derivative is a bit complicated, so I'm suspecting it's not the optimal way.

So, the alternative is proving it directly/Lipschitz condition:
$$\left| {\sin \dfrac{1}{x} + x - \sin \dfrac{1}{y} - y} \right| \le \left| {\sin \dfrac{1}{x} - \sin \dfrac{1}{y}} \right| + \left| {x - y} \right|$$

Now, isn't $\left| {\sin \dfrac{1}{x} - \sin \dfrac{1}{y}} \right|$ bounded by $2$? Because the range of $\sin(x)$ is $[-1,1]$.
So we get:

$$\ldots\le 2 + \left| {x - y} \right|$$

I'm not sure how to proceed from this point, Or maybe it isn't the right one.

Answer 1

Hint (1) a continuous function $f:(a,b)\to\mathbb{R}$ is uniformly continuous iff

$\lim_{x\downarrow a}f(x),\lim_{x\uparrow b}f(x)$ exists.

$\lim_{x\downarrow a}f(x)$ means $x>a$ and approaching to $a$ or we read $x$ decreasing to $a$.

$\lim_{x\uparrow b}f(x)$ means $x<b$ and $x$ is approaching to $b$ so you can say $x$ increasing to $b$.

(2) for a continuous $f$ If $f'(x)$ is bounded then also $f$ is uniformly continuous

Answer 2

The function is asymptotically $x^{3/2}$ so it cannot be uniformly continuous.

To show it explicitly, fix $\delta>0$. Note the estimate (from the Mean Value Theorem) $$ |(x+\delta)^{3/2}-x^{3/2}|\geq\frac{3\delta}2\,\sqrt x. $$ Also, for $x>0$, $$ \sqrt x\,\sin\frac1x\geq-\frac{\sqrt x}{x}=-\frac1{\sqrt x}. $$ Then, for each $k\in\mathbb N$ big enough, $$ |f(k+\delta)-f(k)|=\sqrt{k+\delta}\,\sin\frac1{k+\delta}+(k+\delta)^{3/2}-\sqrt k\,\sin\frac1k-k^{3/2}\\ \geq-\frac1{\sqrt{k+\delta}}-\frac1{\sqrt k}+(k+\delta)^{3/2}-k^{3/2}\\ \geq-\frac1{\sqrt{k+\delta}}-\frac1{\sqrt k}+\frac{3\delta}2\,\sqrt k\geq\delta\sqrt k. $$ So $f$ is not uniformly continuous.