Let ` f ` be a continuous function on `R` such that `f (1/(4n))=sin e^n/(e^(n^2))+n^2/(n^2+1)` Then the value of `f(0) ` is

Options

A. 1
B. 0
C. `-1`
D. `1/6`

Answer

To find the value of \( f(0) \) given the function \( f \) defined by \( f\left(\frac{1}{4n}\right) = \frac{\sin(e^n)}{e^{n^2}} + \frac{n^2}{n^2 + 1} \), we will proceed step by step.

### Step 1: Understand the continuity of \( f \)
Since \( f \) is a continuous function on \( \mathbb{R} \), we can find \( f(0) \) using the limit:
\[ f(0) = \lim_{x \to 0} f(x) \]
We will evaluate this limit as \( n \to \infty \) where \( x = \frac{1}{4n} \).

**Hint:** Use the property of continuity to relate \( f(0) \) to the limit of \( f(x) \) as \( x \) approaches 0.

### Step 2: Substitute \( x = \frac{1}{4n} \)
As \( n \to \infty \), \( \frac{1}{4n} \to 0 \). Thus, we can express \( f(0) \) as:
\[ f(0) = \lim_{n \to \infty} f\left(\frac{1}{4n}\right) \]

**Hint:** Identify how \( f\left(\frac{1}{4n}\right) \) is defined in the problem.

### Step 3: Evaluate the limit
Substituting the definition of \( f \):
\[ f(0) = \lim_{n \to \infty} \left( \frac{\sin(e^n)}{e^{n^2}} + \frac{n^2}{n^2 + 1} \right) \]

**Hint:** Break down the limit into two parts to evaluate them separately.

### Step 4: Evaluate the first term
For the first term:
\[ \lim_{n \to \infty} \frac{\sin(e^n)}{e^{n^2}} \]
Since \( \sin(e^n) \) oscillates between -1 and 1, we have:
\[ -\frac{1}{e^{n^2}} \leq \frac{\sin(e^n)}{e^{n^2}} \leq \frac{1}{e^{n^2}} \]
As \( n \to \infty \), both bounds approach 0. By the Squeeze Theorem:
\[ \lim_{n \to \infty} \frac{\sin(e^n)}{e^{n^2}} = 0 \]

**Hint:** Use the Squeeze Theorem for limits involving oscillating functions.

### Step 5: Evaluate the second term
Now, evaluate the second term:
\[ \lim_{n \to \infty} \frac{n^2}{n^2 + 1} \]
This can be simplified:
\[ \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^2}} = 1 \]

**Hint:** Simplify the expression by dividing the numerator and denominator by \( n^2 \).

### Step 6: Combine the results
Now, combining the results from the two limits:
\[ f(0) = 0 + 1 = 1 \]

### Final Answer
Thus, the value of \( f(0) \) is:
\[ \boxed{1} \]