The limit of (x^2sin (1/x))/(sin x) as x approaches 0 is

Write the ratio as $\dfrac{x}{\sin x}\cdot x\sin\dfrac1x$. The first factor tends to $1$. The second satisfies $\left|x\sin\dfrac1x\right|\le|x|\to0$ by the sandwich theorem, because $\sin\dfrac1x$ is bounded between $-1$ and $1$ although it oscillates. So the limit is $0$. The statement that it does not exist confuses the oscillation of $\sin(1/x)$ with the behaviour of the product, which is squeezed to $0$.