Explanation

(b)
We have,
$f(x) = \left\{ \begin{array}{r}
\frac{2^{x} - 1}{\sqrt{1 + x} - 1},\ - 1 \leq x < \infty,\ \ x \neq 0 \\
k,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0
\end{array} \right.\$
Since, $f(x)$ is continuous everywhere
$\therefore\ \ \ \ \lim_{x \rightarrow 0^{-}}f(x) = f(0)$ ...(i)
Now, $\lim_{x \rightarrow 0^{-}}f(x) = \lim_{h \rightarrow 0}\frac{2^{(0 - h)} - 1}{\sqrt{1 + (0 - h)} - 1}$
$= \lim_{h \rightarrow 0}\frac{2^{- h} - 1}{\sqrt{1 - h} - 1}$
$= \lim_{h \rightarrow 0}\frac{- 2^{- h}\log_{e}2}{\frac{- 1}{2\sqrt{1 - h}}}$ [by L’ Hospital’s rule]
$= 2\ \lim_{h \rightarrow 0}\ 2^{- h}\log_{e}{2\sqrt{1 - h}}$
$= 2\log_{e}2$
From Eq. (i),
$f(0) = 2\log_{e}{2 = \log_{e}4}$