Let f be a continuous function on R such that `f(1//2^n) =(sin e^n) e^(-n^2)+(2n^2)/(n^2+1)` .Then the value of f(0) is

Options

A. 1
B. `1//2`
C. 2
D. none of these

Answer

To find the value of \( f(0) \) for the given continuous function \( f \) defined by

\[ f\left(\frac{1}{2^n}\right) = \sin(e^n)e^{-n^2} + \frac{2n^2}{n^2 + 1}, \]

we will evaluate the limit as \( n \) approaches infinity.

### Step-by-step Solution:

1. **Understanding the Limit**:
Since \( f \) is continuous on \( \mathbb{R} \), we can find \( f(0) \) by taking the limit of \( f\left(\frac{1}{2^n}\right) \) as \( n \to \infty \). As \( n \to \infty \), \( \frac{1}{2^n} \to 0 \).

\[ f(0) = \lim_{n \to \infty} f\left(\frac{1}{2^n}\right) \]

2. **Substituting the Expression**:
Substitute the expression for \( f\left(\frac{1}{2^n}\right) \):

\[ f(0) = \lim_{n \to \infty} \left( \sin(e^n)e^{-n^2} + \frac{2n^2}{n^2 + 1} \right) \]

3. **Evaluating the First Term**:
For the first term \( \sin(e^n)e^{-n^2} \):
- As \( n \to \infty \), \( e^n \to \infty \) and \( \sin(e^n) \) oscillates between -1 and 1.
- However, \( e^{-n^2} \) approaches 0 much faster than \( \sin(e^n) \) can grow, so:

\[ \lim_{n \to \infty} \sin(e^n)e^{-n^2} = 0 \]

4. **Evaluating the Second Term**:
Now consider the second term \( \frac{2n^2}{n^2 + 1} \):

\[ \lim_{n \to \infty} \frac{2n^2}{n^2 + 1} = \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n^2}} = \frac{2}{1 + 0} = 2 \]

5. **Combining the Results**:
Now, combine the results of both terms:

\[ f(0) = 0 + 2 = 2 \]

### Final Answer:
Thus, the value of \( f(0) \) is

\[ \boxed{2} \]