Function is differentiable at x=0?

[frankpupu]

Homework Statement

define a function f:R->R is given by f(x)=(|x|^a)*(sin(1/x)) if x is not equal 0 , and f(x)=0 if x=o . and a>0. for which values of a is f differentiable at x=o

2. The attempt at a solution
obviously, a=2 is one of the solution , but how about other values? can someone can show me the prove for other value of a that this function is differentiable at x=0?

 

[Hurkyl]

Nope! But we'll help you work through the problem.

You seem to already understand a=2 case. How does the proof for that work? In particular, how is a used?

 

[frankpupu]

Nope! But we'll help you work through the problem.

You seem to already understand a=2 case. How does the proof for that work? In particular, how is a used?

for the a=2 case, i use the definition to prove it ,which when cancel the h in both the numerator and denominator then still an h left s.t. hsin(1/h) the limit exists. should it be all the a>1 is the answer?

 

[Hurkyl]

That would be my first conjecture.

Just to check, have you actually proven the derivative doesn't exist for a=1 or a<1? Or did you stop once you realized the algebraic manipulation you wanted to do assumes a>1?

 

[frankpupu]

That would be my first conjecture.

Just to check, have you actually proven the derivative doesn't exist for a=1 or a<1? Or did you stop once you realized the algebraic manipulation you wanted to do assumes a>1?

the question just considered the a>0 , for 0<a<1 we have (sin(1/h))/(|h|^(1-a)) for h>0 then the denominator is 0 not defined and the same for h<0, for a=1 it is not continuouse then for a>1 we have (|h|^(a-1))sin(1/h)=(|h|^a)(hsin(1/h)) then both (|h|^a) (hsin(1/h))are differentiable then the function is differentiable and the same as the negative one


does it make sense??