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A limit from derivative

Calculus (Integrals, Series)
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Papapetros Vaggelis
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A limit from derivative

#1

Post by Papapetros Vaggelis » Tue Nov 10, 2015 9:10 pm

Let \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}\) be a differentiable function at \(\displaystyle{x_0=0}\) and

\(\displaystyle{f(0)=0}\) . If \(\displaystyle{k\in\mathbb{N}}\), then evaluate the limit :



$$\lim_{x\to 0}\dfrac{1}{x}\,\left[f(x)+f\,\left(\dfrac{x}{2}\right)+...+f\,\left(\dfrac{x}{k}\right)\right]$$
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Tolaso J Kos
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Re: A limit from derivative

#2

Post by Tolaso J Kos » Tue Nov 10, 2015 9:11 pm

Here is a solution. Let \( \mathcal{H}_k \) denote the \(k \) - th harmonic number. We are evaluating the limit according to the limit derivative definition:

$$\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}= f'(0)$$

So , according to the above limit we split the limit accordingly:


\( \begin{align*}
\lim_{x\rightarrow 0}\left [ \frac{f(x)-f(0)}{x}+\frac{1}{2}\cdot \frac{f\left (\frac{x}{2} \right )-f(0)}{\frac{x}{2}-0} +\cdots + \frac{1}{k}\cdot \frac{f\left ( \frac{x}{k} \right )-f(0)}{\frac{x}{k}-0}\right ] &=f'(0)\left ( 1+ \frac{1}{2} + \cdots + \frac{1}{k}\right ) \\
&= \mathcal{H}_k f'(0)
\end{align*} \)
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