Let `lim_(n->oo)((x^2+2x+3+sinpix)^n-1)/((x^2+2x+3+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 finite number of points

Answer

To solve the limit problem, we will analyze the expression step by step.

Given:
\[ \lim_{n \to \infty} \frac{(x^2 + 2x + 3 + \sin(\pi x))^n - 1}{(x^2 + 2x + 3 + \sin(\pi x))^n + 1} \]

### Step 1: Define the function
Let:
\[ f(x) = x^2 + 2x + 3 + \sin(\pi x) \]
Then the limit can be rewritten as:
\[ \lim_{n \to \infty} \frac{f(x)^n - 1}{f(x)^n + 1} \]

### Step 2: Analyze the function \( f(x) \)
We need to determine the range of \( f(x) \):
- The term \( \sin(\pi x) \) oscillates between -1 and 1.
- Therefore, the minimum value of \( f(x) \) occurs when \( \sin(\pi x) = -1 \):
\[ f(x) \geq x^2 + 2x + 3 - 1 = x^2 + 2x + 2 \]
- The minimum value of \( x^2 + 2x + 2 \) can be found by completing the square:
\[ x^2 + 2x + 2 = (x + 1)^2 + 1 \geq 1 \]
- Hence, \( f(x) \geq 1 \).

### Step 3: Evaluate the limit
Since \( f(x) \geq 1 \) for all \( x \), we can analyze the limit:
\[ \lim_{n \to \infty} \frac{f(x)^n - 1}{f(x)^n + 1} \]
As \( n \to \infty \):
- If \( f(x) > 1 \), then \( f(x)^n \to \infty \).
- Thus, we can simplify:
\[ \frac{f(x)^n - 1}{f(x)^n + 1} \approx \frac{f(x)^n}{f(x)^n} = 1 \]

### Step 4: Conclusion
Therefore, we conclude:
\[ \lim_{n \to \infty} \frac{(x^2 + 2x + 3 + \sin(\pi x))^n - 1}{(x^2 + 2x + 3 + \sin(\pi x))^n + 1} = 1 \]
for all \( x \in \mathbb{R} \).

### Final Answer
The limit evaluates to:
\[ 1 \]