Question

Statement $I:$ If $f\left(x\right)=\{\begin{array}{cc}x,& \text{if}x\text{is rational}\\ 1-x,& \text{if}x\text{is irrational}\end{array},$ then $\underset{x\to 1/2}{\text{lim}}f\left(x\right)$ does not exist.

Statement $II:$ $x\to \frac{1}{2}$ can be a rational or an irrational value

• A. Both $I$ and $II$ are individually true and $II$ is the correct explanation of $I$
• B. Both $I$ and $II$ are individually true but $II$ is not the correct explanation of $I$
• C. $I$ is true but $II$ is false
• D. $I$ is false but $II$ is true

Answer 1

The correct option is D $I$ is false but $II$ is true

Obviously, statement $II$ is true, as on the number line, immediate neighbourhood of $\frac{1}{2}$ is either rational or irrational, but this does not stop $f\left(x\right)$ to have a limit at $x=\frac{1}{2}.$

As $f\left(\frac{1}{2}\right)=\frac{1}{2},$
$f\left(\frac{1}{2^{+}}\right)=\underset{x\to 1/2^{+}}{\text{lim}}x=\frac{1}{2}$ (if $\frac{1}{2^{+}}$ is rational ) (or)

$\underset{x\to 1/2^{+}}{\text{lim}}(1-x)=1-\frac{1}{2}=\frac{1}{2}$ (if $\frac{1}{2^{+}}$ is irrational)

Hence, $\underset{x\to 1/2^{+}}{\text{lim}}f\left(x\right)=\frac{1}{2}$

With similar argument, we can prove that

$\underset{x\to 1/2^{-}}{\text{lim}}f\left(x\right)=\frac{1}{2}.$

Hence, limit of function exists at $x=\frac{1}{2}.$