The Limits of Sine

$\begin{aligned} \lim_{x \to 0} \frac {x^2 \sin \frac 1x}{\sin x} & = \lim_{x \to 0} \sin \frac 1x \cdot \frac {x^2}{\sin x} \end{aligned}$

We note that $-1 \le \sin \frac 1x \le 1$ is bounded and $\displaystyle \lim_{x \to 0} \dfrac {x^2}{\sin x} = 0$ by L’Hôpital's rule , therefore, $\displaystyle \lim_{x \to 0} \frac {x^2 \sin \frac 1x}{\sin x} = \boxed{0}$

$\displaystyle \lim_{x \to 0} \frac{x^2 sin(1/x)}{sinx} \\ =\displaystyle \lim_{x \to 0} \frac{x}{sinx} \times \displaystyle \lim_{x \to 0} x sin(1/x) \\ =1 \times \displaystyle \lim_{x \to 0} x sin(1/x)$

As $x \rightarrow 0$ then $sin(1/x) \to sin(\infty)$ which will oscillates between $[-1,1]$ .So this will be finite.Finite value multiplied with zero equals to zero.

So, the limit will be zero.