Is the Function B(x)= xsin(1/x) Differentiable at x=0?

[sl2382]

Homework Statement

B(x)= xsin(1/x) when x is not equal to 0

= 0 when x is equal to 0

Determine if the function is differentiable at 0

Homework Equations

The Attempt at a Solution

I get B'(x)= sin(1/x)+cos(1/x)*(-1/x) but really do not know what should be done next.. I think for B'(x) x cannot be 0, but isn't the discontinuity removed as the function is defined to be 0 at x=0? ...

 

[QuarkCharmer]

So it's a piecewise function right?

What does the function have to be in order for it to be differentiable? Check with the definition of the derivative.

 

[sl2382]

So it's a piecewise function right?

What does the function have to be in order for it to be differentiable? Check with the definition of the derivative.

I think the requirements are: 1)the derivative exists at a point 2)limits approaching from both sides of that point are the same ?

So, the derivative does not exist at 0, BUT isn't it defined at 0 ? Does that mean the derivative actually exists and the function can be differentiate?

 

[Dick]

Write the derivative as a difference quotient. f'(0) should be lim h->0 (f(h)-f(0))/h. Pick a specific sequence approaching 0, say h_n=1/(pi*n/2) for n an integer. So h_n->0 as n->infinity. Is there a limit? It's actually pretty helpful to sketch a graph of the function.