If `f` is a positive function such that `f(x+T)=f(x)(T>0), AA x in R,` then `lim_(n->oo)n((f(x+T)+2f(x+2T)+.......+nf(x+n T))/(f(x+T)+4f(x+4T)+.......+n^2f(x+n^2T)})=`

Options

A. 2
B. `2/3`
C. `3/2`
D. none of these

Answer

To solve the limit problem, we start with the given function \( f \) which is periodic with period \( T \), meaning \( f(x + T) = f(x) \) for all \( x \in \mathbb{R} \). We want to evaluate the limit:

\[ \lim_{n \to \infty} n \left( \frac{f(x+T) + 2f(x+2T) + \ldots + nf(x+nT)}{f(x+T) + 4f(x+4T) + \ldots + n^2f(x+n^2T)} \right) \]

### Step 1: Rewrite the sums using the periodic property of \( f \)

Since \( f(x + kT) = f(x) \) for any integer \( k \), we can simplify both the numerator and the denominator:

- The numerator becomes:
\[ f(x+T) + 2f(x+2T) + \ldots + nf(x+nT) = f(x) (1 + 2 + \ldots + n) = f(x) \cdot \frac{n(n+1)}{2} \]

- The denominator becomes:
\[ f(x+T) + 4f(x+4T) + \ldots + n^2f(x+n^2T) = f(x) (1 + 4 + 9 + \ldots + n^2) = f(x) \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 2: Substitute the sums into the limit expression

Now we substitute these results back into the limit expression:

\[ \lim_{n \to \infty} n \left( \frac{f(x) \cdot \frac{n(n+1)}{2}}{f(x) \cdot \frac{n(n+1)(2n+1)}{6}} \right) \]

### Step 3: Simplify the limit expression

We can cancel \( f(x) \) (since \( f(x) > 0 \)) and simplify:

\[ = \lim_{n \to \infty} n \cdot \frac{6 \cdot n(n+1)}{2 \cdot n(n+1)(2n+1)} \]

This simplifies to:

\[ = \lim_{n \to \infty} \frac{6n}{2(2n+1)} = \lim_{n \to \infty} \frac{3n}{2n + 1} \]

### Step 4: Evaluate the limit

Now, we can evaluate the limit:

\[ = \lim_{n \to \infty} \frac{3n}{2n + 1} = \lim_{n \to \infty} \frac{3}{2 + \frac{1}{n}} = \frac{3}{2} \]

### Final Answer

Thus, the final answer is:

\[ \frac{3}{2} \]