# Differentiable once not twice

### Differentiable once not twice

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

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, $$f(x)= x^3 sin(1/x^2)$$ also goes to 0. Since f(0)= 0, the function is continous at x= 0 so may be differentiable there.

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

In order to find the derivative at x= 0 we need to use the definition of the derivative: $$\lim_{h\to 0}\frac{f(h)- f(0)}{h}=$$$$\lim_{h\to 0} \frac{h^3 sin(1/h^2)- 0}{h}=$$$$\lim_{h\to 0} h^2 sin(1/h^2)=$$$$\lim_{h\to 0} \frac{sin(1/h^2)}{1/h^1}$$. To take that limit, let $$u= 1/h^2$$ so the limit becomes $$\lim_{u\to \infty} \frac{sin(u)}{u}$$. 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 $$\lim_{x\to 0} f'(x)$$ does not exist, f' is not continuous at x=0 so is not differentiable there- f''(0) does not exist.
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