Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$? WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:
$$
 \lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) =  \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
=  \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1  \\
= 1?
$$
I wonder where I'm wrong...

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