Continuity of $\sqrt{x}\sin{\frac{1}{x}}$ I want to determine what values of f are continuous then show it, where 
$f(x) = \begin{cases} 
      \sqrt{x}\sin{\frac{1}{x}} & x \ne 0 \\
      0 & x=0 
   \end{cases}
$
Given that $\sqrt{x}$ is not defined for values of $x \lt 0$, f is not continuous in that range.
Also, at $x = 0$, there has to be a value arbitrarily close to x, a such that $a \lt 0$ hence $f(a)$ is undefined so $f$ is not continuous at $x = 0$
I see no reason for $f$ to not be continuous at $x \gt 0$, so then I want to show that that is true.
To prove this, I started by fixing $a \in \Bbb{R}$ then letting $\epsilon \gt 0$ and then choose $\delta$ based on how the rest of proof finishes. Either way, I can suppose that $0 \lt |x-a| \lt \delta$ and since $x, a \gt 0$, we can get $x + a \lt \delta$ or something similar, again, based on later parts of the proof.
Then I've tried a variety of things to manipulate $|f(x)-f(a)|$, the most hopeful being $|f(x)-f(a)|^2 \le x + a + 2\sqrt{a}\sqrt{x} \lt \delta + 2\sqrt{a}\sqrt{x}$ from before.
My issue is I do not know how to finish this proof or even if this is a good method.
Thanks in advance!
edit: This is my idea based on a comment. 
Its known that $\sqrt{x}, \sin{\left(x\right)}, and \frac{1}{x}$ are continuous. so  $ \sin{\left(\frac{1}{x}\right)}$ is continuous and hence $\sqrt{x}\sin{\frac{1}{x}}$ is continuous as it's the product of continous functions.
Is this correct?
 A: If you want to regard $f$ as a function $\Bbb{R} \to \Bbb{R}$, then it needs to be modified; for instance like
\begin{align}
f(x) &=
\begin{cases}
\sqrt{|x|}\sin\left(\frac{1}{x} \right) & \text{if $x \neq 0$}\\
0 & \text{if $x = 0$}
\end{cases}
\end{align}
In this case, since the absolute value $|\cdot|$, square root $\sqrt{\cdot}$, taking reciprocals, products etc are all continuous functions, it follows that $f$ is continuous away from the origin. To prove continuity at the origin, note that for all $x \in \Bbb{R}\setminus\{0\}$, we have
\begin{align}
|f(x) - f(0)| &= \left|\sqrt{|x|}\sin\left(\frac{1}{x} \right) - 0 \right| \\
& \leq \sqrt{|x|} & \tag{since $-1 \leq \sin(\cdot) \leq 1$}
\end{align}
Clearly, $\lim\limits_{x \to 0} \sqrt{|x|} = 0$; hence by the squeeze theorem, it follows that $\lim \limits_{x \to 0}f(x)$ exists and equals $0 = f(0)$. Hence, $f$ is continuous at the origin as well.

If you don't want to modify the function, then of course you have to consider it as a map $f: [0, \infty) \to \Bbb{R}$, and a very similar arugment shows that it is still continuous on all of $[0, \infty)$.
