# differentiable function

• December 2nd 2006, 09:06 AM
dave007rules
differentiable function
Hello. I have this problem,
If r>0 is a rational number, let f: R to R be defined by f(x)=(abs(x)^r)*sin(1/x) for x not equalling zero and f(0)=0. Determine the values of r for which f'(0) (first derivative) exists.

I could really really use some help. Thank you.
• December 2nd 2006, 02:32 PM
ThePerfectHacker
What we have is,
$f(x)=\left\{ \begin{array}{c}|x|^{a/b}\sin (1/x) \mbox{ for }x\not = 0\\ 0 \mbox{ for }x=0\end{array}\right\}$

If the function fails to be continous at $x=0$ (it is continous everywhere else because the product of two continous functions at a point is continous at that point).

So what is the limit,
$\lim_{x\to 0}|x|^{a/b}\sin \frac{1}{x}$ equal to?
If, $a=0$ then,
$|x|^{a/b}=1$.
And we have discontinuity be the limit
$\lim_{x\to 0}\sin \frac{1}{x}$
Does not exist.
So we need to consider the cases when $a\not = 0$.

We require the limit from the right to conincide with the limit from the left. Since $|x|$ is always positive and $\sin \frac{1}{x}$ does change from positive and negative. We can only have for limit existence to be 0. Which is persicely what the functional value at the point is, $f(0)=0$.
Thus, we can consider cases when the right-handed limits are zero.
In that case,
$\lim_{x\to 0^+} x^{a/b}\sin \frac{1}{x}$
We can write,
$\lim_{x\to 0^+} x^{a/b-1} x\sin \frac{1}{x}$
(But that limit does not exist, ever.
If you appraoch it a even or odd multiplies of $\pi$)*

*)I did not prove that, but it makes sense to me.
• December 2nd 2006, 02:40 PM
dave007rules
What are the actual values of a/b though that make it continuous?
• December 2nd 2006, 02:54 PM
ThePerfectHacker
Quote:

Originally Posted by dave007rules
What are the actual values of a/b though that make it continuous?

Look at what I posted (assuming my last conjecture statement was true). None.
• December 2nd 2006, 03:26 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
Look at what I posted (assuming my last conjecture statement was true). None.

I think that you need to rethink that last statement.
$\begin{array}{l}
- 1 \le \sin \left( {\frac{1}{x}} \right) \le 1 \\
- \left| x \right|^r \le \left| x \right|^r \sin \left( {\frac{1}{x}} \right) \le \left| x \right|^r ,\quad r \ge 1 \\
\end{array}$
• December 2nd 2006, 03:29 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
I think that you need to rethink that last statement.
$\begin{array}{l}
- 1 \le \sin \left( {\frac{1}{x}} \right) \le 1 \\
- \left| x \right|^r \le \left| x \right|^r \sin \left( {\frac{1}{x}} \right) \le \left| x \right|^r ,\quad r \ge 1 \\
\end{array}$

See, I was thinking what happens when we approach zero as,
$\frac{1}{\pi (2k)}$
And,
$\frac{1}{\pi (2k+1)}$
• December 2nd 2006, 10:00 PM
CaptainBlack
Quote:

Originally Posted by dave007rules
Hello. I have this problem,
If r>0 is a rational number, let f: R to R be defined by f(x)=(abs(x)^r)*sin(1/x) for x not equalling zero and f(0)=0. Determine the values of r for which f'(0) (first derivative) exists.

I could really really use some help. Thank you.

Other posters have shown that the function is continuous at x=0 as long as
differentiability at x=0.

As x approaches zero sin(1/x) has zeros closer and closer together with the
interval between them going to zero, so there are always points inside and
interval [-epsilon, epsilon] such that:

[f(0)-f(x)]/x=0, x!=0,

Also if the derivative of d(0) of |x|^r is not zero, there are other points in the same interval for which:

[f(0)-f(x)]/x ~= d(0)

which means that f(x) is not differentiable at 0. I will leave it to the reader
to show that of d(0)=0 then f is differentiable at 0, and its derivative is 0.

That the derivative of |x|^r is zero at x=0 means that r>1.

RonL