Show that $f^{(k)}(x_0)=0$ for all $k=0,1,2,...,n$ if and only if $g(x)= {f(x)}/{(x-x_0)^{n+1}}$ is continuous at $x_0$.

Question

Question: Suppose $f:[a,b]\rightarrow R$ has $n+1$ continuous derivatives and $x_0\in(a,b)$. Show that $f^{(k)}(x_0)=0$ for all $k=0,1,2,...,n$ if and only if $g(x)= \frac {f(x)}{(x-x_0)^{n+1}}$ is continuous at $x_0$.

Attempt: 1. $f^{(k)}(x_0)=0 \longrightarrow g(x)= \frac {f(x)}{(x-x_0)^{n+1}}$ continuous at $x_0$.

$f^{(k)}(x_0)=0 \rightarrow g^{(k)}(x_0)(n+1)!(x_0-x_0) = 0.$ (*)It implies that $g^{(k)}(x_0)$ exists. Therefore, $g$ is continuous at $x_0$ by the theorem that if $f$ is differentiable at $c$, it is continuous at $c$.

2. $g(x)= \frac {f(x)}{(x-x_0)^{n+1}}$ continuous at $x_0$$ \longrightarrow f^{(k)} (x_0)=0$

Could you tell me how to prove the converse, and is (*) correct?

Thank you in advance.

Answer 1

Assume the right sided, then \begin{align*} \lim_{x\rightarrow x_{0}}g(x)=g(x_{0}) \end{align*} by assumption, and hence \begin{align*} \lim_{x\rightarrow x_{0}}g(x)(x-x_{0})^{n}=g(x_{0})(x_{0}-x_{0})^{n}=0, \end{align*} but \begin{align*} \lim_{x\rightarrow x_{0}}g(x)(x-x_{0})^{n}=\lim_{x\rightarrow x_{0}}\dfrac{f(x)}{x-x_{0}}=0. \end{align*} If we express $f$ as the Taylor form that \begin{align*} & f(x)\\ &=f(x_{0})+f'(x_{0})(x-x_{0})+\cdots+\dfrac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+\dfrac{1}{(n+1)!}f^{(n+1)}(\xi)(x-x_{0})^{n+1}, \end{align*} since $g$ is continuous, we have first that $f(x_{0})=0$, so \begin{align*} f(x)&=f'(x_{0})(x-x_{0})+\dfrac{1}{2!}f^{(2)}(x_{0})(x-x_{0})^{2}\cdots+\dfrac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}\\ &~~~~+\dfrac{1}{(n+1)!}f^{(n+1)}(\xi)(x-x_{0})^{n+1}, \end{align*} and hence \begin{align*} \dfrac{f(x)}{x-x_{0}}&=f'(x_{0})+\dfrac{1}{2!}f^{(2)}(x_{0})(x-x_{0})+\cdots+\dfrac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n-1}\\ &~~~~+\dfrac{1}{(n+1)!}f^{(n+1)}(x_{0})(x-x_{0})^{n}, \end{align*} and so \begin{align*} \lim_{x\rightarrow x_{0}}\dfrac{f(x)}{x-x_{0}}&=f'(x_{0})+\lim_{x\rightarrow x_{0}}\left(\dfrac{1}{2!}f^{(2)}(x_{0})(x-x_{0})+\cdots+\dfrac{1}{(n+1)!}f^{(n+1)}(\xi)(x-x_{0})^{n}\right)\\ &=f'(x_{0}), \end{align*} so $f'(x_{0})=0$, one can perform similar procedure until $n$. For example, for $k=2$: We already have $f'(x_{0})=0$, so back to the Taylor formula, we have \begin{align*} f(x)&=\dfrac{1}{2!}f''(x_{0})(x-x_{0})^{2}+\cdots+\dfrac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n}+\dfrac{1}{(n+1)!}f^{(n+1)}(\xi)(x-x_{0})^{n+1}, \end{align*} and that \begin{align*} \lim_{x\rightarrow x_{0}}\dfrac{f(x)}{(x-x_{0})^{2}}&=f^{(2)}(x_{0})\\ &~~~~+\lim_{x\rightarrow x_{0}}\left(\cdots+\dfrac{1}{n!}f^{(n)}(x_{0})(x-x_{0})^{n-2}+\dfrac{1}{(n+1)!}f^{(n+1)}(\xi)(x-x_{0})^{n-1}\right)\\ &=f^{(2)}(x_{0}). \end{align*} On the other hand, \begin{align*} \lim_{x\rightarrow x_{0}}\dfrac{f(x)}{(x-x_{0})^{2}}=\lim_{x\rightarrow x_{0}}g(x)(x-x_{0})^{n-1}=g(x_{0})(x_{0}-x_{0})^{n-1}=0. \end{align*}

Answer 2

You just need Taylor Theorem for the direct version. Thus if $f^{(k)} (x_0)=0$ for all $k=0,1,2,\dots, n$ then we have via Taylor's theorem $$f(x) =\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(c)$$ for some $c$ between $x_0$ and $x$. And therefore $$g(x) =\frac{f(x)} {(x-x_0)^{n+1}}=\frac{f^{(n+1)}(c)}{(n+1)!}$$ Taking limits as $x\to x_0$ and noting that $f^{(n+1)}$ is continuous we get $$\lim_{x\to x_{0}}g(x)=\frac{f^{(n+1)}(x_0)}{(n+1)!}$$ For the converse we are given that the limit of $g(x) $ as $x\to x_{0}$ exists and thus we have $$\lim_{x\to x_0}\frac{f(x)}{(x-x_0)^{n+1}}=L$$ and then we have $$f(x_0)=\lim_{x\to x_{0}}f(x)=\lim_{x\to x_0}\frac{f(x)}{(x-x_0)^{n+1}}\cdot(x-x_0)^{n+1}=L\cdot 0=0$$ Next we have $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)}{x-x_0}=\lim_{x\to x_0}\frac{f(x)}{(x-x_0)^{n+1}}\cdot(x-x_0)^{n}=L\cdot 0=0$$ To proceed further in general it is best to apply L'Hospital's Rule. On the contrary let $k$ be the least value for which $f^{(k)} (x_0)\neq 0$ and $f^{(k-1)}(x_0)=f^{(k-2)}(x_0)=\dots=f(x_0)=0$. Applying L'Hospital's Rule $k$ times on RHS of the equation $$L=\lim_{x\to x_{0}}\frac{f(x)}{(x-x_0)^{n+1}}$$ we get $$L=\lim_{x\to x_0}\frac{f^{(k)}(x)}{(n+1)n(n-1)\dots(n+1-k+1)(x-x_{0})^{n+1-k}}$$ and this leads to a contradiction as numerator tends to a non-zero value and denominator tends to $0$ making the original limit infinite.