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. not unique
B. 1
C. data sufficient to find f(0)
D. data insufficient to find f(0)

Answer

To find the value of \( f(0) \) for the function defined as

\[ f\left(\frac{1}{4n}\right) = \frac{\sin(e^n)}{e^{n^2}} + \frac{n^2}{n^2 + 1}, \]

we will follow these steps:

### Step 1: Understand the limit condition
We need to find \( f(0) \). Since \( f \) is continuous, we can find \( f(0) \) by evaluating the limit of \( f\left(\frac{1}{4n}\right) \) as \( n \to \infty \).

### Step 2: Set up the limit
As \( n \to \infty \), \( \frac{1}{4n} \to 0 \). Thus, we need to evaluate:

\[ \lim_{n \to \infty} f\left(\frac{1}{4n}\right) = \lim_{n \to \infty} \left( \frac{\sin(e^n)}{e^{n^2}} + \frac{n^2}{n^2 + 1} \right). \]

### Step 3: Evaluate the first term
Consider the first term \( \frac{\sin(e^n)}{e^{n^2}} \). As \( n \to \infty \), \( e^n \) approaches infinity, and thus \( \sin(e^n) \) oscillates between -1 and 1. Therefore, we can say:

\[ -1 \leq \sin(e^n) \leq 1. \]

This implies:

\[ -\frac{1}{e^{n^2}} \leq \frac{\sin(e^n)}{e^{n^2}} \leq \frac{1}{e^{n^2}}. \]

As \( n \to \infty \), both \( -\frac{1}{e^{n^2}} \) and \( \frac{1}{e^{n^2}} \) approach 0. By the Squeeze Theorem:

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

### Step 4: Evaluate the second term
Now consider the second term \( \frac{n^2}{n^2 + 1} \):

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

### Step 5: Combine the limits
Putting both limits together, we have:

\[ \lim_{n \to \infty} f\left(\frac{1}{4n}\right) = 0 + 1 = 1. \]

### Conclusion
Since \( f \) is continuous, we conclude that:

\[ f(0) = 1. \]

Thus, the value of \( f(0) \) is \( \boxed{1} \).

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