Let `f:R rarr R`be differentiable at x = 0. If f(0) = 0 and `f'(0)=2`, then the value of
`lim_(xrarr0)(1)/(x)[f(x)+f(2x)+f(3x)+….+f(2015x)]` is

Options

A. 2015
B. 0
C. `2015xx2016`
D. `2015xx2014`

Answer

To solve the limit problem, we start by analyzing the given function \( f \) which is differentiable at \( x = 0 \) with \( f(0) = 0 \) and \( f'(0) = 2 \). We need to evaluate the limit:

\[ L = \lim_{x \to 0} \frac{1}{x} [f(x) + f(2x) + f(3x) + \ldots + f(2015x)] \]

### Step 1: Rewrite the expression inside the limit
We can express the limit as:

\[ L = \lim_{x \to 0} \frac{f(x) + f(2x) + f(3x) + \ldots + f(2015x)}{x} \]

### Step 2: Apply L'Hôpital's Rule
Since both the numerator and denominator approach \( 0 \) as \( x \to 0 \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

- The derivative of the denominator \( x \) is \( 1 \).
- The derivative of the numerator \( f(x) + f(2x) + f(3x) + \ldots + f(2015x) \) requires the chain rule:

\[ \frac{d}{dx}[f(kx)] = f'(kx) \cdot k \]

Thus, the derivative of the numerator is:

\[ f'(x) + 2f'(2x) + 3f'(3x) + \ldots + 2015f'(2015x) \]

### Step 3: Substitute back into the limit
Now we can rewrite the limit using the derivatives:

\[ L = \lim_{x \to 0} \left( f'(x) + 2f'(2x) + 3f'(3x) + \ldots + 2015f'(2015x) \right) \]

### Step 4: Evaluate the limit as \( x \to 0 \)
As \( x \to 0 \), \( f'(kx) \) approaches \( f'(0) \) for all \( k \). Since \( f'(0) = 2 \), we have:

\[ L = f'(0) \left( 1 + 2 + 3 + \ldots + 2015 \right) \]

### Step 5: Calculate the sum of the series
The sum of the first \( n \) natural numbers is given by the formula:

\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]

For \( n = 2015 \):

\[ 1 + 2 + 3 + \ldots + 2015 = \frac{2015 \times 2016}{2} \]

### Step 6: Substitute back into the limit
Now substituting back into our limit expression:

\[ L = 2 \cdot \frac{2015 \times 2016}{2} = 2015 \times 2016 \]

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

\[ \boxed{2015 \times 2016} \]