Differentiable once not twice

[Guest]

Hi, I'm really struggling with this problem I can differentiate twice but don't understand the x=0 part. Any help is appreciated.

[Guest]

I had responded to this on another board but if you haven't seen it:
First, note that sine is always between -1 and 1 so as x goes to 0, [tex]f(x)= x^3 sin(1/x^2)[/tex] also goes to 0. Since f(0)= 0, the function is continous at x= 0 so may be differentiable there.

For x not 0, [tex]f(x)= x^3 sin(1/x^2)[/tex]. Using the product rule and the chain rule, [tex]f'(x)= 3x^2 sin(1/x^2)+ x^3cos(1/x^2)(-2/x^{3})= 3x^2 sin(1/x^2)+ cos(1/x^2)[/tex]. That does NOT have a limit as x goes to 0 but it is NOT necessary that "[tex]f'(0)= \lim_{x\to 0} f(x)[/tex]".

In order to find the derivative at x= 0 we need to use the definition of the derivative: [tex]\lim_{h\to 0}\frac{f(h)- f(0)}{h}=[/tex][tex]\lim_{h\to 0} \frac{h^3 sin(1/h^2)- 0}{h}=[/tex][tex]\lim_{h\to 0} h^2 sin(1/h^2)=[/tex][tex]\lim_{h\to 0} \frac{sin(1/h^2)}{1/h^1}[/tex]. To take that limit, let [tex]u= 1/h^2[/tex] so the limit becomes [tex]\lim_{u\to \infty} \frac{sin(u)}{u}[/tex]. Again, sin(x) is always between -1 and 1. Since the denominator is going too infinity, that limit is 0. The derivative at x= 0 exists and is f'(0)= 0.

Since [tex]\lim_{x\to 0} f'(x)[/tex] does not exist, f' is not continuous at x=0 so is not differentiable there- f''(0) does not exist.