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...
 A: $$\lim_{h\to0}\frac{\sin h}h=1$$ 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, $\lim_{h\to0}h=0$
$$\implies  \lim_{h\to0}h\cdot \sin \frac1h =0 $$
A: $$\lim_{x \to 0} \left(x\cdot \sin\left(\dfrac{1}{x}\right)\right) = \lim_{\large\color{blue}{\bf x\to 0}} 
 \left(\frac{\sin\left(\dfrac{1}{x}\right)}{\frac 1x}\right) = 
 \lim_{\large\color{blue}{\bf x \to \pm\infty}} \left(\frac{\sin x}{x}\right) = 0 \neq 1$$
A: Since $-1\leq \sin(1/x)\leq 1$, we have $-x\leq x \sin(1/x)\leq x $. Hence, by squeze theorem we have $\lim_{x\rightarrow 0} x \sin(1/x)=0$.
A: 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.
$$\lim_{x\to x_o}f(x)=0\;\;,\;\;|g(x)|\le M\;,\;\forall\,x\in\text{I}_0\;,\;\;\text{for some constant}\;M\in\Bbb R\;,\;\text{then}$$
$$\lim_{x\to x_0}f(x)g(x)=0$$
A: In the second fraction, you're confusing $\lim_{x\to 0}$ and $\lim_{x \to\infty}$.
A: $$
\lim_{x\to0} \frac{\sin \frac1x}{\frac1x} = \lim_{u\to\pm\infty} \frac{\sin u}{u} = 0\qquad\text{where }u=\frac1x.
$$
A: Hint:  Use the definition of derivative.
