If $f(x) = \begin{cases} x\sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right)$ is equal to

Let $f: R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$. Then $\lim_{x \to \infty} \frac{f(2x)}{f(x)} = $

$\lim _{x \rightarrow 2}\left[\left(x^2-4 x+4\right) \cos \left(\frac{2}{x-2}\right)+\frac{x^2-4}{x^3-2 x-4}\right]=$

If $f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q} \\ -x, & \text{if } x \in \mathbb{Q}^c \end{cases}$,then $\lim_{x \to 0} f(x)$ is