Let `f(x)=lim_(n to oo) ((2 sin x)^(2n))/(3^(n)-(2 cos x)^(2n)), n in Z`. Then

Options

A. at `x=n pm(pi)/(6)`, f(x) is discontinuous
B. `f((pi)/(3))=1`
C. f(0)=0
D. all of the above

Explanation

To solve the problem, we need to evaluate the limit given by the function \( f(x) = \lim_{n \to \infty} \frac{(2 \sin x)^{2n}}{3^n - (2 \cos x)^{2n}} \).

### Step 1: Analyze the limit expression

We start with the expression:

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

### Step 2: Identify the dominant terms as \( n \to \infty \)

We need to consider the behavior of the numerator and denominator as \( n \) approaches infinity. The numerator is \( (2 \sin x)^{2n} \) and the denominator is \( 3^n - (2 \cos x)^{2n} \).

1. If \( |2 \sin x| < 3 \), then \( (2 \sin x)^{2n} \) will approach 0 as \( n \to \infty \).
2. If \( |2 \sin x| = 3 \), then both terms in the denominator will grow at the same rate.
3. If \( |2 \sin x| > 3 \), then the numerator will dominate.

### Step 3: Evaluate specific cases

#### Case 1: \( |2 \sin x| < 3 \)

If \( |2 \sin x| < 3 \), we have:

\[ f(x) = \lim_{n \to \infty} \frac{(2 \sin x)^{2n}}{3^n} = 0 \]

#### Case 2: \( |2 \sin x| = 3 \)

This occurs when \( \sin x = \pm \frac{3}{2} \), which is not possible since \( \sin x \) must be in the range \([-1, 1]\).

#### Case 3: \( |2 \sin x| > 3 \)

This occurs when \( \sin x > \frac{3}{2} \) or \( \sin x < -\frac{3}{2} \), which is also not possible.

### Step 4: Identify discontinuities

Now, we need to check specific values of \( x \):

1. **At \( x = 0 \)**:
\[ f(0) = \lim_{n \to \infty} \frac{(2 \cdot 0)^{2n}}{3^n - (2 \cdot 1)^{2n}} = \frac{0}{3^n - 4^n} = 0 \]

2. **At \( x = \frac{\pi}{3} \)**:
\[ f\left(\frac{\pi}{3}\right) = \lim_{n \to \infty} \frac{(2 \cdot \frac{\sqrt{3}}{2})^{2n}}{3^n - (2 \cdot \frac{1}{2})^{2n}} = \lim_{n \to \infty} \frac{3^n}{3^n - 1} = 1 \]

3. **At \( x = \frac{\pi}{6} \)**:
\[ f\left(\frac{\pi}{6}\right) = \lim_{n \to \infty} \frac{(2 \cdot \frac{1}{2})^{2n}}{3^n - (2 \cdot \frac{\sqrt{3}}{2})^{2n}} = \lim_{n \to \infty} \frac{1^{2n}}{3^n - 3^n} \text{ (undefined)} \]

### Conclusion

From the evaluations:

- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{3}\right) = 1 \)
- \( f\left(\frac{\pi}{6}\right) \) is undefined, indicating a discontinuity.

Thus, the function is discontinuous at \( x = \frac{\pi}{6} \), \( f\left(\frac{\pi}{3}\right) = 1 \), and \( f(0) = 0 \).

### Final Answer

The correct option is **D: All of the above**.