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

\quad \lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty}\left(1+\cos ^{2 m} n ! \,\,\,x \,\,\,\pi\right) is equal to ( x is rational):

Option: 1

2


Option: 2

1


Option: 3

0


Option: 4

3


Answers (1)

best_answer

We know that |\cos \theta| \leq 1 for all \theta. Also |\cos n ! \pi x|<1, if x is irrational.

Hence \lim _{m \rightarrow 0} \lim _{n \rightarrow 0}\left(1+\cos ^{2 m} n ! \pi x\right)=1 and (i.e. \left.x=\frac{p}{q} p, q \in \mid\right) . n ! x \piis an integral multiple of \pi. Hence \cos n ! \pi x=1 or  -1 and \cos ^{2 m} n ! x \pi=1.

Hence \lim _{m \rightarrow 0} \lim _{n \rightarrow 0}\left(1+\cos ^{2 m} n ! \pi x\right)=2

Hence (A) is the correct answer.

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Sayak

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