The point of discontinuity of `f(x)=underset(n to oo)lim ((2 sin x)^(2n))/(3^(n)-(2 cos x)^(2n)), n in Z` is

Options

A. (a) ` at x = npi +-(pi)/6 f(x) is discontinuous`
B. `npi +-(pi)/3p n epsilonI`
C. `n pi +-(pi)/6`
D. None of these

Answer

To find the points of discontinuity of the function

\[ f(x) = \lim_{n \to \infty} \frac{(2 \sin x)^{2n}}{3^n - (2 \cos x)^{2n}}, \]

we need to analyze the behavior of the function as \( n \) approaches infinity.

### Step 1: Identify the condition for discontinuity
The function \( f(x) \) will be discontinuous if the denominator approaches zero. Therefore, we need to set the denominator equal to zero:

\[ 3^n - (2 \cos x)^{2n} = 0. \]

### Step 2: Rearranging the equation
From the equation above, we can rearrange it to:

\[ 3^n = (2 \cos x)^{2n}. \]

### Step 3: Expressing the equation in terms of powers
We can rewrite the right-hand side:

\[ 3^n = 2^{2n} (\cos x)^{2n}. \]

### Step 4: Taking the ratio of the bases
Now we can express this as:

\[ \left(\frac{3}{4 \cos^2 x}\right)^n = 1. \]

### Step 5: Analyzing the limit
For this equation to hold true as \( n \to \infty \), the base must equal 1, which gives us:

\[ \frac{3}{4 \cos^2 x} = 1. \]

### Step 6: Solving for \( \cos^2 x \)
From the equation above, we can solve for \( \cos^2 x \):

\[ 4 \cos^2 x = 3 \implies \cos^2 x = \frac{3}{4}. \]

### Step 7: Finding \( \cos x \)
Taking the square root gives us:

\[ \cos x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}. \]

### Step 8: Finding the general solution for \( x \)
The general solutions for \( \cos x = \frac{\sqrt{3}}{2} \) and \( \cos x = -\frac{\sqrt{3}}{2} \) are:

1. For \( \cos x = \frac{\sqrt{3}}{2} \):
\[ x = 2n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}. \]

2. For \( \cos x = -\frac{\sqrt{3}}{2} \):
\[ x = (2n + 1)\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}. \]

### Conclusion
Thus, the points of discontinuity of the function \( f(x) \) are given by:

\[ x = n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}. \]

---