If $f(x) = \frac{3x - 4}{2x - 3}$, then $f(f(f(x)))$ will be

Let $f(x) = 1 - x$, $g(x) = \frac{1}{1 - x}$, and $h(x) = \frac{1}{x}$ be three functions, for $x \neq 0, 1$. If a function $F(x)$ satisfies $f(F(h(x))) = g(x)$, then which of the following is true?

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$, then $\lim _{x \rightarrow 0} g(f(x))$

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are two functions such that $g \circ f: A \rightarrow C$ is a bijection,then which one of the following is always true?

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping, then:

If function $f: R \rightarrow R$ is defined by $f(x) = (3 - x^5)^{\frac{1}{5}}$,then $(f \circ f)(x) = $ . . . . . . .