Explanation

(d)
LHL$= \ \lim_{x \rightarrow 2^{-}}\frac{5}{\sqrt{2} - \sqrt{x}}$
$= \ \lim_{h \rightarrow 0}{\frac{5}{(\sqrt{2} - \sqrt{2 - h})} \times \frac{(\sqrt{2} + \sqrt{2 - h})}{(\sqrt{2} + \sqrt{2 - h})}}$
$= \lim_{h \rightarrow 0}\ \frac{5(\sqrt{2} + \sqrt{2 - h})}{2 - 2 + h} = \infty$
RHL$= \lim_{x \rightarrow 2^{+}}\frac{5}{\sqrt{2} - \sqrt{x}}$
$= \ \lim_{h \rightarrow 0}{\frac{5}{(\sqrt{2} - \sqrt{2 + h})} \times \frac{(\sqrt{2} + \sqrt{2 + h})}{(\sqrt{2} + \sqrt{2 + h})}}$
$= \lim_{h \rightarrow 0}\ \frac{5(\sqrt{2} + \sqrt{2 + h})}{2 - 2 - h} = - \infty$
$\therefore$ LHL$\neq$RHL
Hence, limit does not exist.