WBJEE · Maths · Limits
Let \(f: R \rightarrow R\) be differentiable at \(x=0\). If f(0)=0 and f'(0)=2, then the value of \[
\left.\lim _{x \rightarrow 0} \frac{1}{x} \mid f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)\right] \text{ is}
\]
- A 2015
- B 0
- C \(2015 \times 2016\)
- D \(2015 \times 2014\)
Answer & Solution
Correct Answer
(C) \(2015 \times 2016\)
Step-by-step Solution
Detailed explanation
Given, \(f(0)=0\) and \(f^{\prime}(0)=2\) \(\therefore \lim _{x \rightarrow 0} \frac{1}{x}[f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]\)…
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