If `f(x)` is continuous in `[0, 1]` and `f((1)/(3))=12`, then the value of `lim_(nrarroo)f((sqrtn)/(3sqrtn+1))` is equal to

Options

A. 2
B. 3
C. 12
D. None of these

Answer

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

\[ \lim_{n \to \infty} f\left(\frac{\sqrt{n}}{3\sqrt{n} + 1}\right) \]

### Step 1: Simplifying the Argument of the Function
First, we simplify the expression inside the function \( f \):

\[ \frac{\sqrt{n}}{3\sqrt{n} + 1} \]

As \( n \) approaches infinity, both the numerator and the denominator approach infinity. We can factor out \( \sqrt{n} \) from the denominator:

\[ \frac{\sqrt{n}}{3\sqrt{n} + 1} = \frac{\sqrt{n}}{\sqrt{n}(3 + \frac{1}{\sqrt{n}})} = \frac{1}{3 + \frac{1}{\sqrt{n}}} \]

### Step 2: Taking the Limit
Now, we take the limit as \( n \) approaches infinity:

\[ \lim_{n \to \infty} \frac{1}{3 + \frac{1}{\sqrt{n}}} = \frac{1}{3 + 0} = \frac{1}{3} \]

### Step 3: Applying Continuity of \( f \)
Since \( f(x) \) is continuous on the interval \([0, 1]\), we can substitute the limit we found into the function:

\[ \lim_{n \to \infty} f\left(\frac{\sqrt{n}}{3\sqrt{n} + 1}\right) = f\left(\frac{1}{3}\right) \]

### Step 4: Using the Given Information
From the problem, we know that \( f\left(\frac{1}{3}\right) = 12 \).

### Final Result
Thus, the value of the limit is:

\[ \lim_{n \to \infty} f\left(\frac{\sqrt{n}}{3\sqrt{n} + 1}\right) = 12 \]

### Conclusion
The final answer is:

\[ \boxed{12} \]