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let $f$ be a differentiable function at $x = 0$ and $f(0) = 0.$ For each $n\in N$, calculate the following limit: $I = \lim_{x \rightarrow > 0} {\frac{1}{x}\left( f(x)+f(\frac{x}{2})+\cdots+f(\frac{x}{n})\right)}$

I know $f'(0)=\lim_{h \rightarrow 0}{\frac{f(h)}{h}}$, so does $I=f'(0)\frac{n(n+1)}{2}?$

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    $\begingroup$ 1+1/2+...+1/n ≠ n(n+1)/2 $\endgroup$
    – user600016
    Dec 5, 2021 at 7:41

1 Answer 1

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Taking limits for individual terms and adding them up we see that the limit is $f'(0)[1+\frac 1 2+\frac 1n+\cdots+\frac 1n]$.

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  • $\begingroup$ So my answer right $I=f'(0)\frac{n(n+1)}{2}$ $\endgroup$
    – tompi2394
    Dec 5, 2021 at 7:33
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    $\begingroup$ @tompi2394 How does summing an harmonic progression give you $\frac{n(n+1)}{2}$? . The answer is just $f'(0)H_{n}$. $H_{n}$ is nth harmonic number $\endgroup$ Dec 5, 2021 at 7:40
  • $\begingroup$ @Mr.GandalfSauron, Oh, Thank you, my bad! $\endgroup$
    – tompi2394
    Dec 5, 2021 at 13:31

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