calculus - Evaluate $lim_{x to infty} sqrt{x} sin(frac{1}{x})$

I am trying to evaluate $\lim_{x \to \infty} \sqrt{x} \sin(\frac{1}{x})$. I have been taught that L'Hopital's Rule is only valid for fractions $\frac{f(x)}{g(x)}$ which have the form where $f(x) = g(x) = 0$ or where $g(x) = \pm \infty$ and $f(x)$ is anything.

Right away I notice that this limit evaluates to $\infty \cdot 0$. I need to put this in either the $\frac{0}{0}$ or the $\frac{\text{anything}}{\pm \infty}$ form to use L'Hopital's rule.

So I write $\sqrt{x} \sin(\frac{1}{x}) = \frac{\sin(\frac{1}{x})}{\frac{1}{\sqrt{x}}}$ to get it in the $\frac{\text{anything}}{\pm \infty}$ form. Executing L'Hopital's Rule I find $\lim_{x \to \infty} \sqrt{x} \sin(\frac{1}{x}) = \lim_{x \to \infty} \frac{2\cos(\frac{1}{x})}{\sqrt{x}} = \frac{0}{\infty}$ but this is yet another undetermined form is it not? It seems fairly obvious to me that the limit here is zero but also that we have an undetermined form. Do I have to keep doing rounds of L'Hopital's Rule?