Let `f(x)=lim_(nrarroo)((x^(2)+2x+4+sinpix^(n))-1)/((x^(2)+2x+4+sinpix^(n))+1)`, then

Options

A. `f(x)` is continuous and differentiable for all `x in R`.
B. `f(x)` is continuous but not differentiable for all `x in R`.
C. `f(x)` is discontinuous at infinite number of points.
D. f(x) is discontinuous at two points.

Answer

To solve the problem, we need to evaluate the limit:

\[ f(x) = \lim_{n \to \infty} \frac{x^2 + 2x + 4 + \sin(\pi x^n) - 1}{x^2 + 2x + 4 + \sin(\pi x^n) + 1} \]

### Step 1: Simplify the expression
First, we can rewrite the limit expression:

\[ f(x) = \lim_{n \to \infty} \frac{(x^2 + 2x + 4 + \sin(\pi x^n)) - 1}{(x^2 + 2x + 4 + \sin(\pi x^n)) + 1} \]

Let \( \alpha = x^2 + 2x + 4 + \sin(\pi x^n) \). The limit now becomes:

\[ f(x) = \lim_{n \to \infty} \frac{\alpha - 1}{\alpha + 1} \]

### Step 2: Analyze the behavior of \(\sin(\pi x^n)\)
As \( n \to \infty \), the term \( \sin(\pi x^n) \) oscillates between -1 and 1 depending on the value of \( x \). Therefore, we need to analyze the behavior of \( \alpha \):

\[ \alpha = x^2 + 2x + 4 + \sin(\pi x^n) \]

### Step 3: Find the minimum value of \(\alpha\)
The quadratic part \( x^2 + 2x + 4 \) has a minimum value. To find this minimum, we can complete the square or use calculus:

\[ x^2 + 2x + 4 = (x + 1)^2 + 3 \]

The minimum value occurs at \( x = -1 \):

\[ \text{Minimum value} = 3 \]

Since \( \sin(\pi x^n) \) oscillates between -1 and 1, the minimum value of \( \alpha \) is:

\[ \text{Minimum of } \alpha = 3 - 1 = 2 \]
\[ \text{Maximum of } \alpha = 3 + 1 = 4 \]

Thus, we conclude that \( \alpha \) is always greater than 1:

\[ \alpha > 2 > 1 \]

### Step 4: Evaluate the limit
Since \( \alpha > 1 \), we can evaluate the limit:

\[ f(x) = \lim_{n \to \infty} \frac{\alpha - 1}{\alpha + 1} \]

As \( n \to \infty \), \( \sin(\pi x^n) \) will not affect the limit significantly, and we can consider the dominant terms:

\[ f(x) = \frac{\alpha - 1}{\alpha + 1} \]

### Step 5: Consider the behavior of \( \alpha \)
As \( n \to \infty \), we can say that:

\[ \alpha \to x^2 + 2x + 4 + \text{(oscillating term)} \]

Thus, we can apply L'Hôpital's Rule if needed, but since \( \alpha \) is bounded, we can directly conclude:

\[ f(x) = 1 \]

### Conclusion
Therefore, the final result is:

\[ f(x) = 1 \]