Evaluating $\lim_{x \to 0} \frac{\sin^3(x)\sin(\frac{1}{x})}{x^2}$

Question

Evaluate $$\lim_{x \to 0} \frac{\sin^3(x)\sin(\frac{1}{x})}{x^2}$$

My attempt: $$\lim_{x \to 0} \frac{\sin^3(x)\sin(\frac{1}{x})}{x^2}=\lim_{x \to 0} \frac{\sin^3(x)\sin(\frac{1}{x})}{x^3}\cdot x$$

$$=\lim_{x \to 0} \sin^3(x)\cdot \frac{\sin(\frac{1}{x})}{\frac{1}{x}}\cdot \frac{1}{x^3}$$

Is this the right path? If so, what should I do next?

Answer 1

Notice, $\sin\left(\frac1x\right)\in [-1,1] \ \ \ \forall \ \ x\in \mathbb R$ $$\lim_{x \to 0} \frac{\sin^3(x)\sin(\frac{1}{x})}{x^2}=\lim_{x \to 0} \underbrace{\frac{\sin^3x}{x^3}}_{\to 1}\cdot \underbrace{\sin(\frac{1}{x})}_{-1\le\large \boxed{} \le 1}\cdot \underbrace{x}_{\to 0}$$ $$=0$$

Answer 2

We use the fact that $$|\sin(x)|\le |x|$$ and $$|\sin(\frac 1x)|\le 1$$ to get

$$|\frac{\sin^3(x)\sin(\frac 1x)}{x^2}|\le |x|$$

thus

$$\lim_{x\to 0}\frac{\sin^3(x)\sin(\frac 1x)}{x^2}=0$$