# how many times can we differentiate this function??

• February 5th 2009, 10:11 PM
transgalactic
how many times can we differentiate this function??
how many times can we differentiate this splitted function on point x=0
??
$
f(x) = \{ x^{2n} \sin (\frac{1}{x})
,x \ne 0

$

{0,x=0}

i did one differentiation
what to do next??
$
f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0
$

$
f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0
$
• February 5th 2009, 10:52 PM
Abu-Khalil
You asumme $n=1$ or $x^{2n}$ was an error?

Anyway, knowing that, you may define the derivate function of $f(x)$ as $f'(x)=\begin{cases}2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right), x\not=0\\
0,x=0
\end{cases}$
.

And now you have problems because $f''(x)=\lim_{x\to0}\frac{2x\sin\left(\frac{1}{x}\r ight)-\cos\left(\frac{1}{x}\right)-0}{x-0}=\lim_{x\to0}2\sin\left(\frac{1}{x}\right)+\frac {\cos\left(\frac{1}{x}\right)}{x}$ doesn't converge.
• February 7th 2009, 10:50 PM
transgalactic
what is the correct way of proving it??
• February 8th 2009, 07:02 AM
HallsofIvy
Proving what? That $\lim_{x\rightarrow 0} 2 sin(1/x)+ \frac{cos(1/x)}{x}$ does not exist? That should be obvious: cos(1/x) is always between -1 and 1 and the denominator goes to 0 so that term does not converge. 2 sin(1/x) is always between -2 and 2 so it cannot "offset" the fact that cos(1/x)/x diverges.
• February 8th 2009, 08:30 AM
transgalactic
i got a different result.
i assumed that in point x=0 the function turns to 0 too.
$
f(x) = \ x^{2n} \sin (\frac{1}{x})
$

$
f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})
$

$
f''(x)=\lim_{x->0+}\frac{2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})-0}{x-0}=\\

$

$
f''(x)=\lim_{x->0+}\frac{2nx^{2n-2}\sin (\frac{1}{x})-x^{2n-1}\frac{1}{x^2}\cos(\frac{1}{x})-0}{1}
$

so in the plus side i will get 0
and on the minus side
but i cant keep doing this derivatives till i get two different limits
there must be an easier way
??