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  • If x is a real number in , then the value of <img alt="f(x)=\lim _{m \rightarrow \infty} \lim _{n \rightarr

If x is a real number in [0,1], then the value of f(x)=\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty}\left\{1+\cos ^{2 m}(n ! \pi x)\right\} is given by

Option: 1

2 or 1 according as x is rational or irrational
 


Option: 2

1 or 2 according as x is rational or irrational


Option: 3

1 for all x 

 


Option: 4

2 or 1 for all x


Answers (1)

best_answer

\because 0<\cos ^2(n ! \pi x) \leq 1

If                              \because 0<\cos ^2(n ! \pi x) \leq 1

Then, \begin{aligned} f(x) & =\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty}\left\{1+\left(\cos ^2(n ! \pi x)\right)^m\right\} \\ & =1+0=1 \end{aligned}

When x is irrational

and if \cos ^2(n ! \pi x)=1

\Rightarrow \quad n ! \pi x=r \pi \Rightarrow x=\frac{r}{n !}=\text { rational }

Then, 

\begin{aligned} f(x) & =\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty}\left\{1+\left(\cos ^2 n ! \pi x\right)^m\right\} \\ & =1+1=2 \end{aligned}

Hence, 

f(x)=\left\{\begin{array}{l} 2, x \text { is rational } \\ 1, x \text { is irrational } \end{array}\right.

Posted by

Suraj Bhandari

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