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In a script about the Taylor-Theorem I'm reading I have stumbled across the property that for remainder $R_n(x,x_0): \lim_{x \to x_0} \frac{R_n(x,x_0)}{(x-x_0)^n} = 0$. Since there are no further remarks or explanation, does anybody have any insight or hint on how I can understand this better?

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  • $\begingroup$ This is known as Taylor -Young formula: with the notatiions of asymptotic analysis, $\;R_n(x,x_0)=o\bigl((x-x_0)^n\bigr)$. $\endgroup$ – Bernard Aug 24 '19 at 14:05
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Let $P_n(x)$ be $n$th order Taylor polynomial of $f$ about $x_0$. Then $P_n(x)$ is the only polynomial $P(x)$ of degree smaller than or equal to $n$ such that$$\lim_{x\to x_0}\frac{f(x)-P(x)}{(x-x_0)^n}=0.$$Now, use the fact that $f(x)-P_n(x)$ is precisely the remainder of order $n$.

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    $\begingroup$ How would you prove that theorem? $\endgroup$ – Botond Aug 24 '19 at 13:38
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    $\begingroup$ You will find a proof in Michael Spivak's Calculus, for instance. There, it's the corollary of theorem 3, from chapter 20. $\endgroup$ – José Carlos Santos Aug 24 '19 at 13:50
  • $\begingroup$ Thank you, I found it! (+1) $\endgroup$ – Botond Aug 24 '19 at 13:56

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