# 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.

• (A) Since the limit does not exists so the singularity is NOT removable. – Empty Nov 30 '15 at 8:34
• For $(D)$, I think you made a mistake – 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 $$\lim_{x\to x_0^{-}}$$.