Which of the following function(s) has/have removable discontinuity at $x=1?$ Which of the following function(s) has/have removable discontinuity at $x=1?$
$(A)f(x)=\frac{1}{\ln|x|}\hspace{1cm}(B)f(x)=\frac{x^2-1}{x^3-1}\hspace{1cm}(C)2^{-2^{\frac{1}{1-x}}}\hspace{1cm}(D)\frac{\sqrt{x+1}-\sqrt{2x}}{x^2-x}\hspace{1cm}$

I only know that we have a removable discontinuity if the term that makes the denominator of a rational function equal zero for x = a cancels out under the assumption that x is not equal to a.
I applied this definition and found that $(B)$ is a function with removable discontinuity.
But in the answer,functions $(A),(B),(C)$ are given.I do not know how $(A),(C)$ are functions with removable discontinuity.
I found $\lim_{x\to1^-}\frac{1}{\ln|x|}=\lim_{x\to1^-}\frac{1}{\ln x}=-\infty$
I found $\lim_{x\to1^+}\frac{1}{\ln|x|}=\lim_{x\to1^+}\frac{1}{\ln x}=+\infty$
But i do not know why $x=1$ is a removable discontinuity here.
For the function $(C)$,
I found $\lim_{x\to1^-}2^{-2^{\frac{1}{1-x}}}=0$
I found $\lim_{x\to1^+}2^{-2^{\frac{1}{1-x}}}=1$
But i do not know why $x=1$ is a removable discontinuity here.
For the function $(D)$,
$\frac{\sqrt{x+1}-\sqrt{2x}}{x^2-x}=\frac{\sqrt{x+1}-\sqrt{2x}}{x^2-x}\times\frac{\sqrt{x+1}+\sqrt{2x}}{\sqrt{x+1}+\sqrt{2x}}$
$=\frac{1}{-x(\sqrt{x+1}+\sqrt{2x})}$ is a continuous everywhere function.

Please help me.Thanks.
 A: Old question, but I'll stick to the matter at hand, which seems to be your interpretation of the question.
Continuity of $f(x)$ in point $x_0$  means $\lim_{x\downarrow x_0} f(x) = \lim_{x\uparrow x_0} f(x) = f(x_0)$. The graph converges on a point from two sides on the value of the graph at point $x_0$.
The opposite is discontinuity, where any of these three values are different from the others.
Removable discontinuity means $\lim_{x\downarrow x_0} f(x) = \lim_{x\uparrow x_0} f(x) =$ an actual value, and $f(x)$ is undefined at $x_0$. Hence, if you were to fill in that value at $f(x_0)$, the function would be continuous.
All functions $A, B, C, D$ are undefined at $x = 1$, so you need to check if the upper and lower limit match. If they do, they have a removable discontinuity.
For your deductions, you've already seen that A does not approach a value from either the positive or negative side, and for C you've seen that they do not approach the same value.
N.B. My notation differs from yours, $\lim_{x\downarrow x_0}$ means $\lim_{x\to x_0^{+}}$ and $\lim_{x\uparrow x_0}$ means $\lim_{x\to x_0^{-}}$.
