Let f(x) = mathop {lim }limits_{n to infty } | vedclass

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(D) We analyze the limit $f(x) = \mathop {\lim }\limits_{n 	o \infty } \frac{{{{\left( {4\sin^2 x} ight)}^{n}}}}{{{3^n} - {{\left( {4\cos^2 x} ig

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All the above statements are correct.

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(D) We analyze the limit $f(x) = \mathop {\lim }\limits_{n 	o \infty } \frac{{{{\left( {4\sin^2 x} ight)}^{n}}}}{{{3^n} - {{\left( {4\cos^2 x} ight)}^{n}}}}$.Case $1$: If $|4\sin^2 x| This occurs when $\sin^2 x Case $2$: If $|4\sin^2 x| > 3$ and $|4\cos^2 x| This occurs when $\sin^2 x > \frac{3}{4}$,i.e.,$x \in (m\pi + \frac{\pi}{3}, m\pi + \frac{2\pi}{3})$.Case $3$: If $|4\cos^2 x| > 3$ and $|4\sin^2 x| At $x = m\pi \pm \frac{\pi}{6}$,the denominator approaches $0$ or the terms become undefined,leading to discontinuity. Thus,$f(x)$ is discontinuous at these points. Checking $f(0) = 0$ and $f(\pi/3) = 1$ confirms all statements are correct.

Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

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