Which of the following functions has/have removable discontinuity at x=1?

(A)$\underset{\mathrm{x}\to 1}{\lim }\frac{1}{\log \left|\mathrm{x}\right|}$ does not exists 
(B)$\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\lim }\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)(x^2+x+1)}=\frac{2}{3}$ then f has removable discount at x=1
(C) $\underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ does not exist
(D) $$\begin{gathered} \underset{\mathrm{x}\to 1}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\lim }\frac{x+1-2x}{\left(x^2-x\right)\left(\sqrt{\mathrm{x}+1}-\sqrt{2\mathrm{x}}\right)} \\ =\underset{\mathrm{x}\to 1}{\lim }\frac{1-x}{x\left(x^{ }-1\right)\left(\sqrt{\mathrm{x}+1}+\sqrt{2\mathrm{x}}\right)} \\ =\frac{-1}{2\sqrt{2}} \end{gathered}$$ 
       then f has removable discontinous  at x=1