1. ## Derrivative

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.

2. Originally Posted by johntuan
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.
By definition, $\displaystyle f'(0) = \lim_{x\to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x\to 0} \sin \frac{1}{x}$

But this limit does not exist. Why is that?

3. because if u plug in x the denominator is 0?

4. The limit can exist even if the function is not defined at that point..

..there is another reason..

5. im still not too sure but i think it has something to do with the sin right?

6. 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