If x =u is a point of discontinuity of f(x)=`lim_(n to oo) cos^(2n)x`, then the value of cos u is

Options

A. 0
B. `1//2`
C. `(-1)^n`
D. 1

Answer

To solve the problem, we need to analyze the function \( f(x) = \lim_{n \to \infty} \cos^{2n}(x) \) and determine the points of discontinuity.

### Step-by-Step Solution:

1. **Understanding the Function**:
We start with the function:
\[ f(x) = \lim_{n \to \infty} \cos^{2n}(x) \]
We can rewrite this as:
\[ f(x) = \lim_{n \to \infty} (\cos^2(x))^n \]

2. **Analyzing the Limit**:
The behavior of \( f(x) \) depends on the value of \( \cos^2(x) \):
- If \( \cos^2(x) < 1 \), then \( \lim_{n \to \infty} (\cos^2(x))^n = 0 \).
- If \( \cos^2(x) = 1 \), then \( \lim_{n \to \infty} (\cos^2(x))^n = 1 \).

3. **Finding the Conditions**:
The condition \( \cos^2(x) = 1 \) occurs when:
\[ \cos(x) = \pm 1 \]
This happens when:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]

4. **Identifying Points of Discontinuity**:
At points \( x = n\pi \):
- \( f(n\pi) = 1 \) (since \( \cos(n\pi) = \pm 1 \)).
- For \( x \) not equal to \( n\pi \), \( f(x) = 0 \).

Therefore, \( f(x) \) is discontinuous at \( x = n\pi \) because:
- The left-hand limit \( \lim_{x \to n\pi^-} f(x) = 0 \)
- The right-hand limit \( \lim_{x \to n\pi^+} f(x) = 0 \)
- But \( f(n\pi) = 1 \)

5. **Conclusion**:
Since \( u \) is a point of discontinuity, we have \( u = n\pi \). Now, we need to find \( \cos(u) \):
\[ \cos(u) = \cos(n\pi) \]
The value of \( \cos(n\pi) \) is:
- \( \cos(n\pi) = (-1)^n \)

### Final Answer:
Thus, the value of \( \cos(u) \) is:
\[ \cos(u) = (-1)^n \]