Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?

$$\lim_{h\to0}\frac{\sin h}h=1$$ but $$\lim_{h\to0}\frac{\sin \frac1h}{\frac1h}=\lim_{h\to0}h\cdot \sin \frac1h$$

Now, $-1\le \sin \frac1h\le 1$

$$\implies \left|h\cdot \sin \frac1h\right|\le \left|h\right| $$

Now, $\lim_{h\to0}h=0$

$$\implies \lim_{h\to0}h\cdot \sin \frac1h =0 $$

$$\lim_{x \to 0} \left(x\cdot \sin\left(\dfrac{1}{x}\right)\right) = \lim_{\large\color{blue}{\bf x\to 0}} \left(\frac{\sin\left(\dfrac{1}{x}\right)}{\frac 1x}\right) = \lim_{\large\color{blue}{\bf x \to \pm\infty}} \left(\frac{\sin x}{x}\right) = 0 \neq 1$$

Since $-1\leq \sin(1/x)\leq 1$, we have $-x\leq x \sin(1/x)\leq x $. Hence, by squeze theorem we have $\lim_{x\rightarrow 0} x \sin(1/x)=0$.