If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$,then

If $f(x) = \begin{cases} \sin x, & x \ne n\pi, n \in \mathbb{Z} \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \ne 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x))$ is

If $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$,then $f(f(x))$ is

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$. Then for all values of $x$,the value of $f(g(x))$ is:

If $f(x)$ and $g(x)$ are two functions with $g(x)=x-\frac{1}{x}$ and $f \circ g(x)=x^3-\frac{1}{x^3}$,then $f^{\prime}(x)$ is equal to

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$