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

Let $f(x) = ax + b$ and $g(x) = cx + d$,where $a \ne 0$ and $c \ne 0$. Assume $a = 1$ and $b = 2$. If $(fog)(x) = (gof)(x)$ for all $x$,what can you say about $c$ and $d$?

Given the functions $f: R \rightarrow R$ defined by $f(x) = 2x^2 - 5$ and $g: R \rightarrow R$ defined by $g(x) = \frac{x}{x^2 + 1}$,find the composite function $(g \circ f)(x)$.

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 $f: R \to R$ is defined by $f(x) = (x + 1)^2$ and $g: R \to R$ is defined by $g(x) = x^2 + 1$,then $(fog)(-3)$ equals:

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\begin{cases} \log _e x & , x>0 \\ e^{-x} & , x \leq 0 \end{cases}$ and $g(x)=\begin{cases} x & , x \geq 0 \\ e^{x} & , x < 0 \end{cases}$. Then $gof: R \to R$ is . . . .