Determine whether f'(0) exists.
f(x)={ xsin(1/x) if x does not =0
{
{ 0 if x=0
{
Sorry but its suposed to be a piecewise function.
Actually there is a point which is seen hard
Assume if you put in sin1/x any number , it goes a range if you put it in 0 ''zero '' is criticall 1/x x=0 result of this is the infinitesimal sin inf. is not available but if you think that sin x for every x value goes [-1,1] we all know there is no chance for sinx and if we put ( x=inf. ) it always goes between [-1,1] range but none of them is not taken by it so we account its range ''not available'' this equation helps us generally limit and derivative problems