how many times can we differentiate this splitted function on point x=0

??

[math]

f(x) = \{ x^{2n} \sin (\frac{1}{x})

,x \ne 0

 

 

 

[/math]

on {0,x=0}

I don't see why you shouldn't be able to differentiate it indefinitely.

 

Crudely,

[math]f^{(n)}(x) = \begin{cases}

\frac{d^n}{dx^n} x^{2}\sin(x^{-1}), & x \neq 0 \\

\lim\limits_{x \to 0}\left ( \frac{d^n}{dx^n} x^{2}\sin(x^{-1}) \right ) , & x=0

\end{cases}[/math]

 

Why would there be a problem there?

I don't see why you shouldn't be able to differentiate it indefinitely.

I do!

 

Crudely,

[math]f^{(m)}(x) = \begin{cases}

\frac{d^m}{dx^m} x^{2n}\sin(x^{-1}), & x \neq 0 \\

\lim\limits_{x \to 0}\left ( \frac{d^m}{dx^m} x^{2n}\sin(x^{-1}) \right ) , & x=0

\end{cases}[/math]

 

Why would there be a problem there?

Because you have to show that the limit exists.

 

Hint: Explicitely try n=0, n=1 and n=2 for the 1st derivative, first. You should already see a pattern from there.

Oh damn. I disregarded the n out of laziness, I wasn't expecting it to make a difference for some reason.

 

Well, if you're willing to throw continuity to the wind then you could still go for:

 

[math]

f^{(m)}(x) = \begin{cases}

\frac{d^m}{dx^m} x^{2n}\sin(x^{-1}) & x \neq 0 \\

0 & x=0

\end{cases}

[/math]

Oh damn. I disregarded the n out of laziness, I wasn't expecting it to make a difference for some reason.

 

Well, if you're willing to throw continuity to the wind ...

That's the point of this exercise. You cannot throw continuity to the wind. A pre-condition for a function to be differentiable at some point is that the function must have both a limit and a value at that point, and these two must be equal to one another.