Limits Continuity And Differentiability Question 45
Question: The set of points of discontinuity of the function $ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{(2\sin x)}^{2}}^{n}}{3^{n}-{{(2\cos x)}^{2n}}} $ is given by
Options:
A) R
B) $ { n\pi \pm \frac{\pi }{3},n\in I } $
C) $ { n\pi \pm \frac{\pi }{6},n\in I } $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have, $ f(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{(2\sin x)}^{2n}}}{-{{(2\cos x)}^{2n}}} $
$ =\underset{n\to \infty }{\mathop{\lim }}\frac{{{(2\sin x)}^{2n}}}{{{(\sqrt{3})}^{2n}}-{{(2\cos x)}^{2n}}} $
$ f(x) $ is discontinuous when $ {{(\sqrt{3})}^{2n}}-{{(2\cos x)}^{2n}}=0 $
i.e. $ \cos x=\pm \frac{\sqrt{3}}{2}\Rightarrow x=n\pi \pm \frac{\pi }{6}(n\in I) $
BETA
