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 find the limit:

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

Given that \( f(x) \) is continuous on the interval \([0, 1]\) and \( f\left(\frac{1}{3}\right) = 12 \).

### Step 1: Simplifying the expression inside the limit

We start by simplifying the expression inside the function \( f \):

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

To simplify this, we can divide the numerator and the denominator by \( \sqrt{n} \):

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

### Step 2: Finding the limit of the simplified expression

Now, we need to evaluate the limit of this expression as \( n \) approaches infinity:

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

As \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \). Therefore, we have:

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

### Step 3: Applying the limit to the function \( f \)

Now we substitute this limit back into the function \( f \):

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

### Step 4: Using the given value of \( f\left(\frac{1}{3}\right) \)

From the problem, we know that:

\[ f\left(\frac{1}{3}\right) = 12 \]

### Conclusion

Thus, the value of the limit is:

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

### Final Answer

The value of the limit is \( 12 \).

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