A certain instance of L'Hospital

Allow $f$ be a function specified on an open period $(a,b)$ which has continual by-products of order $1,2, \dots , n-1$, and also there is a factor $c \in (a,b)$ such that $f(c), f^{{1}}(c), \dots , f^{(n-1)}(c)$ are all $0$. Nonetheless $f^{n}(c)$ is not $0$ and also we can think that $f^{n}(x) > 0$ for $x \in (a,b)$, nonetheless no presumptions concerning the connection of $f^{(n)}(x)$ are made. In this instance is it real that $\lim_{x\to c}\frac{f(x)}{(x-c)^n} = \frac{f^{(n)}(c)}{n!}$?

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2019-05-04 17:16:06
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L'Hospital's regulation does not rely on the connection of the by-product. Under your problems, $\lim_{x\to c}f^{(k)}(x)=0$ for $k\le n-1$, so l'Hospital's regulation can be related to offer : $$\lim_{x\to c}\frac{f(x)}{(x-c)^n} =\lim_{x\to c}\frac{f'(x)}{n(x-c)^{n-1}} =\cdots =\lim_{x\to c}\frac{f^{(n-1)}(x)}{n!(x-c)} =\lim_{x\to c}\frac{f^{(n)}(x)}{n!}. $$

However, to insist that $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ calls for that $f^{(n)}$ be continual at $c$. ($n!$ is independent of $x$, so $\lim_{x\to c}\frac{f^{(n)}(x)}{n!}=\frac{f^{(n)}(c)}{n!}$ and also $\lim_{x\to c}f^{(n)}(x)=f^{(n)}(c)$ are equal and also the last is the definition of connection for $f^{(n)}$ at $c$.)

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2019-05-08 01:17:11
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