Consider the function `f(x)=lim_(nto infty) ((1+cosx)^(n)+5lnx)/(2+(1+cosx)^(n))`, then

Options

A. a. `f(x)` is discontinous at positive even multiples of `pi`
B. b. `f(x)` is continuous at positive odd multiples of `pi`
C. c. `f(x)` is discontinuous at positive odd multiples of `(pi)/(2)`
D. d. `f(x)` is continuous at positive even multiples of `(pi)/(2)`

Explanation

To solve the problem, we need to analyze the given function:

\[ f(x) = \lim_{n \to \infty} \frac{(1 + \cos x)^n + 5 \ln x}{2 + (1 + \cos x)^n} \]

### Step 1: Analyze the term \((1 + \cos x)^n\)
As \(n\) approaches infinity, the behavior of \((1 + \cos x)^n\) will depend on the value of \(1 + \cos x\):
- If \(1 + \cos x > 1\), then \((1 + \cos x)^n \to \infty\).
- If \(1 + \cos x = 1\) (which occurs when \(\cos x = 0\), i.e., \(x = \frac{\pi}{2} + k\pi\) for \(k \in \mathbb{Z}\)), then \((1 + \cos x)^n = 1\).
- If \(1 + \cos x < 1\) (which occurs when \(\cos x < -1\), but this is not possible since \(\cos x\) ranges from -1 to 1), this case does not apply.

### Step 2: Evaluate the limit
We will consider two cases based on the value of \(x\):

**Case 1: \(1 + \cos x > 1\) (i.e., \(x \neq \frac{\pi}{2} + k\pi\))**
In this case, as \(n \to \infty\):
\[ f(x) = \lim_{n \to \infty} \frac{(1 + \cos x)^n + 5 \ln x}{2 + (1 + \cos x)^n} \]
Both the numerator and denominator will be dominated by \((1 + \cos x)^n\):
\[ f(x) = \lim_{n \to \infty} \frac{(1 + \cos x)^n}{(1 + \cos x)^n} = 1 \]

**Case 2: \(1 + \cos x = 1\) (i.e., \(x = \frac{\pi}{2} + k\pi\))**
In this case:
\[ f(x) = \lim_{n \to \infty} \frac{1 + 5 \ln x}{2 + 1} = \frac{1 + 5 \ln x}{3} \]

### Final Result
Thus, we can summarize the function \(f(x)\) as follows:
- For \(x \neq \frac{\pi}{2} + k\pi\), \(f(x) = 1\).
- For \(x = \frac{\pi}{2} + k\pi\), \(f(x) = \frac{1 + 5 \ln x}{3}\).