Which of the following function(s) has/have removable discontinuity at $x=1?$

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 $(D)$,
$=\frac{1}{-x(\sqrt{x+1}+\sqrt{2x})}$ is a continuous everywhere function.

Please help me.Thanks.

  • $\begingroup$ (A) Since the limit does not exists so the singularity is NOT removable. $\endgroup$ – Empty Nov 30 '15 at 8:34
  • $\begingroup$ For $(D)$, I think you made a mistake $\endgroup$ – Claude Leibovici Nov 30 '15 at 8:35

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 $\lim_{x\to x_0^{-}}$.

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