Explanation

(b)
Let $f(x) = \frac{g(x)}{h(x)} = \frac{x}{1 + |x|}$
It is clear that $g(x) = x$ and $h(x) = 1 + |x|$ are differentiable on $( - \infty,\ \infty)$ and $( - \infty,\ 0) \cup (0,\ \infty)$ respectively
Thus, $f(x)$ is differentiable on $( - \infty,\ 0) \cup (0,\ \infty)$.Now, we have to check the differentiable at $x = 0$
$\therefore\ \ \lim_{x \rightarrow 0}\frac{f(x) - f(0)}{x - 0} = \lim_{x \rightarrow 0}\frac{\frac{x}{1 + |x|} - 0}{x} = \lim_{x \rightarrow 0}\frac{1}{1 + |x|} = 1$
Hence, $f(x)$ is differntaible on $( - \infty,\ \infty)$