Question

If $f\left(x\right)=\underset{m\to \infty }{lim}\underset{n\to \infty }{lim}(1+{\text{cos}}^{2m}(n!\pi x))$, then which among the following options is/are CORRECT ?

• A. $f\left(x\right)=2,$ If $x$ is irrational
• B. $f\left(x\right)=1,$ If $x$ is rational
• C. $f\left(x\right)=2,$ If $x$ is rational
• D. $f\left(x\right)=1,$ If $x$ is irrational

Answer 1

Answer: (C), (D)

$f\left(x\right)=1,$ If $x$ is irrational

Case$:1$
When $n\to \infty$ and $x$ is rational i.e., $x=\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0,n!x=n!\times \frac{p}{q}$ is integer as $n!$ has factor $q$ when $n\to \infty$
Also, when $n!x$ is integer,
$\cos (n!\pi x)=\pm 1$
$\therefore \underset{m\to \infty }{lim}\underset{n\to \infty }{lim}(1+{\text{cos}}^{2m}(n!\pi x))=1+1=2$

Case$:2$
When $n\to \infty$ and $x$ is irrational
$\cos (n!\pi x)\in (-1,1)$
$\Rightarrow {\cos }^{2m}(n!\pi x)\to 0$ as $m\to \infty$
$\therefore \underset{m\to \infty }{lim}\underset{n\to \infty }{lim}(1+{\text{cos}}^{2m}(n!\pi x))=1+0=1$