Question
Which of the following function(s) has/have removable discontinuity at $$x = 1 \,?$$
A.
$$f\left( x \right) = \frac{1}{{{\text{ln}}\left| x \right|}}$$
B.
$$f\left( x \right) = \frac{1}{{{x^3} - 1}}$$
C.
$$f\left( x \right) = {2^{{2^{\frac{1}{{1 - x}}}}}}$$
D.
$$f\left( x \right) = \frac{{\sqrt {x + 1} - \sqrt {2x} }}{{{x^2} - x}}$$
Answer :
$$f\left( x \right) = \frac{{\sqrt {x + 1} - \sqrt {2x} }}{{{x^2} - x}}$$
Solution :
$$\eqalign{
& \left( {\bf{A}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr
& \left( {\bf{B}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr
& \left( {\bf{C}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = {\text{ does not exist}}{\text{.}} \cr} $$
$$\left( {\bf{D}} \right)\,\,\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \frac{{ - 1}}{{2\sqrt 2 }},$$ therefore $$f\left( x \right)$$ has removable discontinuity at $$x = 1.$$