Explanation

(d)
$\lim_{x \rightarrow 0}\frac{\sqrt{1 - \cos{2x}}}{\sqrt{2}x} = \lim_{x \rightarrow 0}\frac{\sqrt{2\sin^{2}x}}{\sqrt{2}x}$
$= \lim_{x \rightarrow 0}\frac{\sin{|x|}}{x} = f(x)$ [say]
Now, $f(0 + 0) = \lim_{h \rightarrow 0}\frac{|\sin{(0 + h)}|}{0 + h} = 1$
$f(0 - 0) = \lim_{h \rightarrow 0}\frac{|\sin{(0 - h)}|}{- h} = - 1$
$\because\ \ \ \ \ f(0 + 0) \neq f(0 - 0)$
$\therefore$ The limit of function does not exist.