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$\newcommand{\dirichlet}{\mathop{\rm dirichlet}\nolimits}$ I'm trying to find two examples for the following criterias:

  1. A method that is continuous in exactly one point but doesn't have a derivative in that point
  2. A method that is continuous in exactly one point and does have a derivative in that point

After looking deeper at some examples, I found out that $f(x) = x\cdot\dirichlet(x)$ doesn't have a deriviate at $x = 0$ however $f(x) = x^2\cdot \dirichlet(x)$ does have.

I can't understand the difference between the two. Any helps could help.

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1 Answer 1

up vote 2 down vote accepted

First, $\Delta(x)$ is nowhere continuous, e.g. $\lim_{x\to y}\Delta(x)$ does not exist for any $y$. The definition of the derivative is given by a limit, so if you are looking for the derivative of $x^p\Delta(x)$ for $p\geq 1$ at zero, by the definition you have $$ \lim_{x\to 0}\frac{x^p\Delta(x) - 0^p\Delta(0)}{x-0} = \lim_{x\to0}x^{p-1}\Delta(x) $$ and the question now is, when the latter limit does exist. As we discussed already, it does not exist for $p-1=0$ i.e. $p=1$ hence $x\Delta(x)$ does not have a derivative at zero. However, since $$ \lim_{x\to 0}x^{p-1} = 0 $$ for $p>1$ and $\Delta(x)$ is bounded, the limit $\lim_{x\to0}x^{p-1}\Delta(x)$ does exist and equals zero for all $p>1$.

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I'm sorry, what does x^p represent here? –  vondip Jan 19 '13 at 20:44
1  
@vondip: it's a power. Like you used $x$ or $x^2$ - I just gave an answer for a more general case. What I meant is that when the power is $p=1$ then the derivative does not exist since we have just to take a limit of the Dirichlet function alone. However, whenever $p>1$ (in particular, $p=2$) the limit also involves some continuous function (which is $x^{p-1}$) which converges to zero, and that's why the limit does exist. –  Ilya Jan 19 '13 at 21:10

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