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Let $f$ be a function defined on an open interval $(a,b)$ which has continuous derivatives of order $1,2, \dots , n-1$, and there is a point $c \in (a,b)$ such that $f(c), f^{{1}}(c), \dots , f^{(n-1)}(c)$ are all $0$. However $f^{n}(c)$ is not $0$ and we can assume that $f^{n}(x) > 0$ for $x \in (a,b)$, however no assumptions about the continuity of $f^{(n)}(x)$ are made. In this case is it true that $\lim_{x\to c}\frac{f(x)}{(x-c)^n} = \frac{f^{(n)}(c)}{n!}$?

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That's L'Hôpital to you. –  Noldorin Aug 6 '10 at 16:05
    
@Noldorin: that's my preferred spelling too. From MathWorld: Note that l'Hospital's name is commonly seen spelled both "l'Hospital" (e.g., Maurer 1981, p. 426; Arfken 1985, p. 310) and "l'Hôpital" (e.g., Maurer 1981, p. 426; Gray 1997, p. 529), the two being equivalent in French spelling. –  Isaac Aug 6 '10 at 16:21
    
@Isaac: Yeah. The circumflex accent represents a missing 's', and is the more common (less archaic) spelling... It represents pronunciation better for sure. –  Noldorin Aug 6 '10 at 16:24
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L'Hospital's rule does not depend on the continuity of the derivative. Under your conditions, $\lim_{x\to c}f^{(k)}(x)=0$ for $k\le n-1$, so l'Hospital's rule can be applied to give: $$\lim_{x\to c}\frac{f(x)}{(x-c)^n} =\lim_{x\to c}\frac{f'(x)}{n(x-c)^{n-1}} =\cdots =\lim_{x\to c}\frac{f^{(n-1)}(x)}{n!(x-c)} =\lim_{x\to c}\frac{f^{(n)}(x)}{n!}. $$

However, to assert that $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ requires that $f^{(n)}$ be continuous at $c$. ($n!$ is independent of $x$, so $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ and $\lim_{x\to c}f^{(n)}(x)=f^{(n)}(c)$ are equivalent and the latter is the definition of continuity for $f^{(n)}$ at $c$.)

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