Let `f : R rarr R` be a differentiable function at x = 0 satisfying f(0) = 0 and f'(0) = 1, then the value of `lim_(x to 0) (1)/(x) . sum_(n=1)^(oo)(-1)^(n).f((x)/(n))`, is

Options

A. (a)0
B. (b)`-log 2`
C. (c)1
D. (d)e

Answer

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

\[ \lim_{x \to 0} \frac{1}{x} \sum_{n=1}^{\infty} (-1)^n f\left(\frac{x}{n}\right) \]

Given that \( f(0) = 0 \) and \( f'(0) = 1 \), we can use the Taylor series expansion of \( f(x) \) around \( x = 0 \):

\[ f(x) = f(0) + f'(0)x + o(x) = 0 + 1 \cdot x + o(x) = x + o(x) \]

Now substituting \( f\left(\frac{x}{n}\right) \):

\[ f\left(\frac{x}{n}\right) = \frac{x}{n} + o\left(\frac{x}{n}\right) \]

Now substituting this into the summation:

\[ \sum_{n=1}^{\infty} (-1)^n f\left(\frac{x}{n}\right) = \sum_{n=1}^{\infty} (-1)^n \left(\frac{x}{n} + o\left(\frac{x}{n}\right)\right) \]

This can be separated into two parts:

\[ \sum_{n=1}^{\infty} (-1)^n \frac{x}{n} + \sum_{n=1}^{\infty} (-1)^n o\left(\frac{x}{n}\right) \]

The first part can be simplified:

\[ \sum_{n=1}^{\infty} (-1)^n \frac{x}{n} = x \sum_{n=1}^{\infty} \frac{(-1)^n}{n} = -x \log(2) \]

The second part, since \( o\left(\frac{x}{n}\right) \) goes to zero faster than \( \frac{x}{n} \), will also converge to zero as \( x \to 0 \).

Thus, we have:

\[ \sum_{n=1}^{\infty} (-1)^n f\left(\frac{x}{n}\right) \approx -x \log(2) + o(x) \]

Now substituting this back into our limit:

\[ \lim_{x \to 0} \frac{1}{x} \left(-x \log(2) + o(x)\right) = \lim_{x \to 0} \left(-\log(2) + \frac{o(x)}{x}\right) \]

Since \( \frac{o(x)}{x} \to 0 \) as \( x \to 0 \), we find:

\[ \lim_{x \to 0} \frac{1}{x} \sum_{n=1}^{\infty} (-1)^n f\left(\frac{x}{n}\right) = -\log(2) \]

Thus, the final answer is:

\[ \boxed{-\log(2)} \]