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  • <img alt="\mathrm{\lim _{n \rightarrow \infty} \lim _{n \rightarrow \infty}\left(1+\cos ^{2 m}(n ! \pi x)\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20

\mathrm{\lim _{n \rightarrow \infty} \lim _{n \rightarrow \infty}\left(1+\cos ^{2 m}(n ! \pi x)\right)}

 

Option: 1

if \mathrm{x} is rational then the \mathrm{limit=2} and if \mathrm{x} is irrational then the \mathrm{limit=1}


Option: 2

if \mathrm{x} is rational then \mathrm{limit=1} if \mathrm{x} is irrational then \mathrm{limit=2}


Option: 3

limit will be 2 when \mathrm{x} is rational or irrational


Option: 4

None


Answers (1)

best_answer

We know that \mathrm{\mid \cos \Theta \mid \leq 1} for all 0 .

If \mathrm{ |\cos (n ! \pi x)|=p<1, }

\mathrm{\text { limit }=\lim _{n \rightarrow \infty}\left(1+p^{2 m}\right)=1}        .........(1)

                                          \mathrm{as\: p^{2 m} \rightarrow 0 \: when\: m \rightarrow \infty.}
If \mathrm{ |\cos (n ! \pi x)|=1, }

\mathrm{ \operatorname{limit}=\lim _{n \rightarrow \infty}\left(1+1^{2 m}\right)=\lim _{n \rightarrow \infty} 2=2 .}     ............(2)

Now, \mathrm{ I \: cos (n ! n x) \mid=1}

\mathrm{\Rightarrow n ! \pi x=k n}, where \mathrm{k} is an integer

\mathrm{\therefore \quad x=\frac{k}{n !}} which is a rational number.

\mathrm{\therefore from (2), limit =2 \: if \: x} is rational.

Again, \mathrm{|\cos (n ! \pi x)|<1}

\mathrm{\Rightarrow \quad|\cos (n ! \pi x)| \neq 1 for\: all\: n \in N}

\mathrm{\Rightarrow n ! \pi x \neq k \pi for\: all \: n \in N, k \in Z}

\mathrm{\Rightarrow \quad x \neq \frac{k}{n !} for \: all \: n \in N, k \in Z}

\mathrm{\Rightarrow \quad x \: is \: not \: rational, i.e., x\: is\: irrational.}

\mathrm{\therefore \: from\: \: (1), limit =1\: if\: x\: is\: irrational.}

Hence option 1 is correct.




 

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