Limit of $\sqrt x \sin(1/x)$ where $x$ approaches positive infinity

Question

I have a problem as in the title. I cannot use the typical squeeze theorem strick with $-1<\sin(1/x)<1$ since that does not seem to yield anything useful. I also saw a solution that at small values $0<\sin(\theta)<\theta$ but i would like to avoid that since I have not prove that fact really. I was wondering if I could do the following thing:

We assume that the limit does exist: $\lim \sqrt x \sin(1/x)=L$. Then we square it. $\lim x\sin^2(1/x)=L^2$ and use $u=1/x$ substitution to solve it. My only problem is, i am assuming the limit exists, which i cannot do, so i was wondering if there is a way to fix my proof, or if there is an alternative one that is a nice solution.

Answer 1

It's $$\lim_{x\rightarrow0^+}\left(\frac{\sin{x}}{x}\cdot\sqrt{x}\right)=0.$$

Answer 2

Given that $|\sin a)|\le |a|$ we have

$$\left| \sqrt{x}\sin\left(\frac{1}{x}\right) \right| \le \frac{1}{\sqrt{x}}\to 0~~~~as~~~x\to \infty$$

Indeed, $$ |\sin a)| =\left|\int_0^a\cos(t)dt\right|\le |a|$$

Answer 3

write $$\frac{\sin(\frac{1}{x})}{\frac{1}{x}}\cdot \frac{1}{\sqrt{x}}$$

Answer 4

$$\sin\dfrac1x\sim_\infty\dfrac1x,\;\text{so}\quad\sqrt x\sin\dfrac1x\sim_\infty\frac{\sqrt x}x=\frac1{\sqrt x}\to 0.$$