Let f : R → R be differentiable at x = 0 and satisfies f(0) = 0 and f'(0) = 1, then the value of $\lim\limits_{x \rightarrow 0} \frac{1}{x} \sum\limits_{n=1}^{\infty}(-1)^n f\left(\frac{x}{n}\right)$ is
Answer & explanation
Correct answer: option 2
$\lim\limits_{x \rightarrow 0} \frac{1}{x} \sum\limits_{n=1}^{\infty}(-1)^n f\left(\frac{x}{n}\right)$
$=\lim\limits_{x \rightarrow 0} \sum\limits_{n=1}^{\infty}(-1)^n f\left(\frac{x}{n}\right) . \frac{1}{x}$
$=\lim\limits_{x \rightarrow 0} \sum\limits_{n=1}^{\infty}(-1)^n \times \frac{f\left(\frac{x}{n}\right)-f(0)}{\left(\frac{x}{n}\right)} \times \frac{1}{n}$
$=\sum\limits_{n=1}^{\infty} \times \frac{(-1)^n}{n} \lim\limits_{x \rightarrow 0} \frac{f\left(\frac{x}{n}\right)-f(0)}{\frac{x}{n}}$
$=\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n} f'(0)=\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}=-\ln 2$