Question

Let $x$ be an irrational, then $\underset{m\to \infty }{lim}{\underset{n\to \infty }{lim}\left\{\cos (n!\pi x)\right\}}^{2m}$ equals

• A. $0$
• B. $-1$
• C. $1$
• D. $Indeterminate$

Answer 1

Answer: (A)

$0$

According to the problem

For n=1,2,3......put

$F_n\left(x\right)=\underset{x\to \infty }{lim}{\left(\cos n!\pi x\right)}^{2m}$

when n!x is an integer, $F_n\left(x\right)=0$, For all others values of x,$f\left(x\right)=\underset{x\to \infty }{lim}{f}_n\left(x\right)$

Hence the irrational X,$F_n\left(x\right)=0$

for every m;

Hence $F_n\left(x\right)=0$, for rational X, say $X=\frac{p}{q}$,

where P and q are interger, we see that

n!x is an interger if $m\ge q$ therefore that

f(x)=1

$\underset{m\to \infty }{lim}{\underset{n\to \infty }{lim}\left\{\cos (n!\pi x)\right\}}^{2m}$

0 x irrational

1 x rational

Hence the obtained an every where

discontinuous limit function .