# Thread: Where does f'(0) exist?

1. ## Where does f'(0) exist?

Question: If r>0 is rational, let f(x):= x^r sin(1/x) and f(0):= 0. For what values of r does f'(0) exists?

Attempt: So I know this derivative exists when lim(x->0) {[x^r sin(1/x) - 0]/ [x-0]} exists.
So when lim(x->0) [x^(r-1)sin(1/x)] exists. But I'm a little rusty with limits..

I was testing out r<1, r=1, and r>1 but then I realized I didn't know these limits. Help would be lovely, thanks!

2. You got the right idea upto me.
We want that the following limit to exist
$
\lim_{x\to0}\dfrac{f(x)-f(0)}{x-0}=\lim_{x\to0}x^{r-1}\sin(1/x).
$

To compute this lets substitute $y:=1/x$,
then $x\to0$ means $|y|\to\infty$.
Hence the limit turns out to be
$\lim_{|y|\to\infty}\dfrac{\sin(y)}{y^{r-1}}=0$ if $r>1$. (bounded/unbounded monotonic function)