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...

I think you're confusing $\,x\to 0\,$ with $\,x\to\infty\,$ in the case $\,\frac{\sin x}x\,$ . The general case is
Proposition: Assming $\,f,g\,$ are functions defined in a punctured neighbourhood I$_0\;$ of some point $\,x_0\,$ and s.t.
$$\lim_{x\to x_o}f(x)=0\;\;,\;\;|g(x)|\le M\;,\;\forall\,x\in\text{I}_0\;,\;\;\text{for some constant}\;M\in\Bbb R\;,\;\text{then}$$
$$\lim_{x\to x_0}f(x)g(x)=0$$
