If $f(x) = e^{2x}$ and $g(x) = \log \sqrt{x}$ $(x > 0)$,then $fog(x)$ is equal to

Let $f: R \rightarrow R$ be defined as $f(x)=x-1$ and $g: R -\{1,-1\} \rightarrow R$ be defined as $g(x)=\frac{x^{2}}{x^{2}-1}$. Then the function $f \circ g$ is

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 $f(x) = \sin^2 x + \sin^2(x + \frac{\pi}{3}) + \cos x \cos(x + \frac{\pi}{3})$ and $g(\frac{5}{4}) = 1$,then $(g \circ f)(x) = $

The composite mapping $fog$ of the maps $f:R \to R$,$f(x) = \sin x$,and $g:R \to R$,$g(x) = x^2$ is:

If $f:[0,3] \rightarrow [0,3]$ is defined by $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$,then $f(f(x))$ is: