If $f: R \rightarrow R$ is defined by $f(x) = \frac{x}{x^{2}+1}$,find $f(f(2))$.

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

$f(x) = \begin{cases} 3-x, & -1 \leqslant x < 0 \\ 1+\frac{5x}{3}, & -3 \leqslant x \leqslant 2 \end{cases}$ and $g(x) = \begin{cases} -x, & -2 \leqslant x \leqslant 3 \\ x, & 0 \leqslant x \leqslant 1 \end{cases}$. Find the range of $(f \circ g)(x)$.

For every real number $x \neq -1$,let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$,$f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

If $f$ is the greatest integer function defined on $R$ as $f(x) = [x]$ and $g$ is the modulus function defined on $R$ as $g(x) = |x|$,then the value of $(g \circ f)\left(\frac{-5}{3}\right)$ is

If $f: R \rightarrow R$ is defined by $f(x)= \begin{cases} |[x-5]|, & \text{for } x < 5 \\ [|x-5|], & \text{for } x \geq 5 \end{cases}$ Then,$(f \circ f)\left(-\frac{7}{2}\right) = ?$ (here,$[x]$ is the greatest integer function)