Limit of $\sin(x)\cos(1/x)$ as $x\to 0$

I am at the second lesson of my Calculus 1 course. I found below explanation of the limit $\sin(x)\cos(1/x)$ as $x\to 0$. I plan something similar to use as the answer of my homework. My concern is: is this solution of the limit correct? Can we conclude from the table that limit is approaching to $0$ although it is oscillating and changing the sign?

1 Answer

Yes your guess from the table is correct, indeed since $\;\forall \theta\in\mathbb R\;$ $\;-1\le \cos \theta \le 1$, for $x>0$ we have that

$$-\sin x \le \sin x \cdot \cos \left(\frac1x\right)\le \sin x$$

and since $\sin x \to 0^+$ by squeeze theorem the limit is equal to $0$. For $x<0$ we can use a similar argument.