Solve $\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$

$$0\leq \left| \frac{x^2 \text{sin}\frac{1}{x}}{\text{sin} x} \right| \leq \left| \frac{x^2}{\sin x} \right|$$

You can now apply L'Hôpital's rule to $\frac{x^2}{\text{sin}x}$ to show the limit of this as $x \rightarrow 0$ is $0$, then use the squeeze rule.

You can only split if both limits exist. In this case, it's fine. However, l'Hospital doesn't help here. The derivative of $\sin \frac{1}{x}$ wildly diverges so it's useless. What you need to realize is, that $x\sin\frac{1}{x}$ is a product of a bounded function and a function with limit $0$, which tells you about the limit.

The part $\frac{x}{\sin x}$ converges to 1, that's known.

Rewrite the function as $$ \frac{\left(\frac{\sin \frac1x}{\frac1x}\right) }{\left(\frac{\sin x}{x}\right)} $$ Now think about the behavior of the simpler function $g(x)=\sin x/ x$ both as $x\rightarrow 0$ and as $x\rightarrow \pm \infty$.

As $x\rightarrow 0$, $g(x) \rightarrow 1$.

As $x\rightarrow \pm\infty$, $g(x)\rightarrow 0$.

These are precisely the behaviors of the denominator and numerator, respectively, as $x\rightarrow 0$. So the limit of the original expression as $x\rightarrow 0$ is $0/1=0$.