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The set of points of discontinuity of the function
$f(x)=\lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^{n}-(2 \cos x)^{2 n}}$ is given by
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$f(x)=\lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^{n}-(2 \cos x)^{2 n}}$ is given by
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Verified Answer
The correct answer is:
$\left\{\mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathrm{I}\right\}$
We have, $f(x)=\lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{3^{n}-(2 \cos x)^{2 n}}$
$$
=\lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{(\sqrt{3})^{2 n}-(2 \cos x)^{2 n}}
$$
$\mathrm{f}(\mathrm{x})$ is discontinuous when $(\sqrt{3})^{2 n}-(2 \cos x)^{2 n}=0$
i.e. $\cos x=\pm \frac{\sqrt{3}}{2} \Rightarrow x=n \pi \pm \frac{\pi}{6}$
$$
=\lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{(\sqrt{3})^{2 n}-(2 \cos x)^{2 n}}
$$
$\mathrm{f}(\mathrm{x})$ is discontinuous when $(\sqrt{3})^{2 n}-(2 \cos x)^{2 n}=0$
i.e. $\cos x=\pm \frac{\sqrt{3}}{2} \Rightarrow x=n \pi \pm \frac{\pi}{6}$
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