1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.