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Suppose I know that $\Delta(x)$ is a function on $\mathbb{R}$ such that $\Delta(x) = O(\sqrt{x}).$ Suppose that $x$ is a large real number and that $h < x/2.$ Given this, why does the equality
$$\dfrac{x \log{x} + (2\gamma-1)x+\Delta(x)}{h}-\dfrac{(x-h)\log{x-h}+(2\gamma-1)(x-h)+\Delta(x-h)}{h}$$ $$= \log(x) + 2\gamma + O(\frac{h}{x})+O(\frac{x^{1/2}}{h})$$ hold? Here $\gamma$ is Euler's constant.

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    $\begingroup$ $\frac{\Delta(x)-\Delta(x-h)}{h} = O(\frac{x^{1/2}}{h})$ and you are supposed to say $\Delta$ is the error term in the en.wikipedia.org/wiki/… $\endgroup$ – reuns Jul 27 at 22:28
  • $\begingroup$ @reuns The term I was having trouble explaining was $O(h/x),$ not the $O(x^{1/2}/h).$ Do you have a quick way to deduce this error term? $\endgroup$ – inequalitynoob2 Jul 28 at 8:25
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    $\begingroup$ $\log x - \log (x-h) = -\log(1-h/x) = O(h/x)$ $\endgroup$ – reuns Jul 28 at 12:42

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