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I need to prove that $f(x) =\sqrt{x}{\sin \frac1x}$ is uniformly continuous at $(0, \infty )$.

I managed to show that:

$-\sqrt{x} <= f(x) <= \sqrt{x}$,

$\lim_{x\to0}\sqrt{x} = \lim_{x\to0}-\sqrt{x} = 0$,

and so $\lim_{x\to0}f(x) = 0$

meaning that: $\lim_{x\to0^+}f(x) = 0$.

I also showed: $\lim_{x\to\infty}f(x) = 0$.

How should I continue? Will showing the $f$ is continuous at $(0,1)$ be enough?

If so how can I prove that it is continuous at $(0,1)$?

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    I thought equicontinuity only applies to a set of different functions if they are all uniformly continuous with respect to a single $\delta$ for any chosen $\varepsilon$? If so, I don't see how that applies here, as we're only dealing with one function. Anyway, with regards to the continuity, can you use the results that compositions of continuous functions are continuous themselves? – 2017-01-07
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    @Tom Maybe he means uniformly continuos? – 2017-01-07
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    @Thomas That's what I think as well (can't imagine what it would be otherwise), but I wanted to clarify either way. – 2017-01-07
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    Yes that's what I meant, my bad – 2017-01-07
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    I do not believe the function is uniformly continuous. The derivative is unbounded as $x \to 0.$ – 2017-01-07
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    The derivative being bounded is not a necessary condition of uniform continuity, I believe - it is merely sufficient. See $f(x)=\sqrt{x}$, whose derivative is not bounded close to $0$, but it is still unformly continuous on $[0,1]$. – 2017-01-07
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    You have almost solved the question by evaluating the limits as $x\to 0^{+}$ and as $x\to \infty$. Thus the function can be made continuous in $[0,\infty) $ by defining $f(0)=0$. It is now a trivial matter to show that it is uniformly continuous in $[0,\infty)$ and hence in $(0,\infty)$. – 2017-01-07

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We know that $f(x):=\sqrt{x}\sin\left(\frac{1}{x}\right)$ is continuous on the closed and bounded interval $[0,1]$ - this is easy to see, as you have proven that the limit as $x\rightarrow 0$ exists and is equal to $0$, therefore the function can be extended to a continuous function on $[0,1]$ and this well known result still holds.

We now look at the interval $[1,\infty)$.

Calculating the first derivative, we have:

$$f'(x) = \frac{\sin\left(\frac{1}{x}\right)}{2\sqrt{x}}-\frac{\cos\left(\frac{1}{x}\right)}{x^{\frac{3}{2}}}$$

Note that $f'(1)=\frac{\sin(1)-2\cos(1)}{2}\approx -0.1196$ and that $\lim\limits_{x\rightarrow\infty}f'(x)=0$.

Since $f'(x)$ is continuous on $[1,\infty)$, the fact that these two limits are finite implies that there must exist an upper bound on $|f'(x)|$ for $x\in[1,\infty)$.

Since the derivative is bounded, $f(x)$ is Lipschitz-continuous on $[1,\infty)$, which implies uniform continuity.

As $f(x)$ is uniformly continuous on $[0,1]$ and $[1,\infty)$, it is therefore also uniformly continuous on $[0,\infty)$ (see Paramand Singh's excellent answer on this question as to why the uniform continuity of a function on two sets $A$ and $B$ implies the uniform continuity on $A\cup B$ if $A$, $B$ and $A\cup B$ are intervals) and any subset of this set - thus it is certainly uniformly continuous on $(0,\infty)$.

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    I disagree: the fact that $f$ is uniformly continuous on $A \cup B$ iff it is so on $A$ and $B$ needs some justification. Remember that one definition of uniform continuity is: $f$ is u.c. iff $f(x_n) - f(y_n) \to 0$ whenever $x_n - y_n \to 0$. In your answer, imagine $x_n \to 1$ with $(x_n)_n \subset [0,1]$ and $y_n \to 1$ with $(y_n)_n \subset [1, \infty)$. – 2017-01-07
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    @AlexM.: see this http://math.stackexchange.com/a/1862067/72031 for uniform continuity over $A\cup B$. The current answer is fine. – 2017-01-07
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    I think you have over complicated by going to bounded derivative. Since $f(x) \to 0$ as $x\to\infty$ there is an $M>0$ such that $|f(x) - f(y) |<\epsilon$ for $x, y$ greater than or equal to $M$ so $f$ is uniformly continuous in $[M, \infty) $. And clearly it is uniformly continuous on $[0,M]$. – 2017-01-07
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    @ParamanandSingh: I surely won't downvote this answer, but it *would* be fine only with the said addition. Anyway, your link offers the required complement to this answer. – 2017-01-07
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Denote your function as $f(x)$

First $\lim_{x\rightarrow 0} \sqrt{x}sin{\frac{1}{x}}=0$,add a definition when x=$0$ let your function equals $0$ when $x=0$,so your function is uniformly continous on $(0,a]$,$a>0$.

$f^{'}(x)=\frac{1}{2\sqrt{x}}sin{\frac{1}{x}}-cos{\frac{1}{x}}x^{-\frac{3}{2}}$ is bounded on $[a,\infty)$ with upper bound $|f^{'}(x)|\le \frac{1}{2\sqrt{a}}+a^{-\frac{3}{2}}$,so f continous uniformly on $R_{+}$