# Derivative of piecewise function with $\sin\frac{1}{x}$ term

I was going through my calculus book, and I am not sure I understand this part

$$f(x) = \begin{cases} \frac{x^2}{4}+x^4\sin(\frac{1}{x}) &\text{if x\neq0 } \\ 0 &\text{if x=0 } \end{cases}$$

$$f'(x) = \begin{cases} \frac{x}{2}-x^2\cos(\frac{1}{x})+4x^3\sin(\frac{1}{x}) &\text{if x\neq0 } \\ 0 &\text{if x=0 } \end{cases}$$

$$f''(x) = \begin{cases} \frac{1}{2}+12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x}) &\text{if x\neq0 } \\ \frac{1}{2} &\text{if x=0 } \end{cases}$$

So, I know that when I have piecewise function, I need to look at left and right limit, but I don't see why second part in second derivative is $$\frac{1}{2}$$, or rather why does the $$\sin\frac{1}{x}$$ term go to 0?

• It should be noted that the $\sin \frac{1}{x}$ term does not "go to $0$", and the result shows that while the second derivative $f"(0)$ exists, the second derivative is not continuous at $x=0$. Dec 15, 2018 at 4:06

If $$x\neq0$$, then\begin{align}\frac{f'(x)-f'(0)}x&=\frac{\frac x2-x^2\cos\left(\frac1x\right)+4x^3\sin\left(\frac1x\right)}x\\&=\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\end{align}and therefore\begin{align}f''(0)&=\lim_{x\to0}\frac12-x\cos\left(\frac1x\right)+4x^2\sin\left(\frac1x\right)\\&=\frac12.\end{align}
$$f''(0)=\lim_{x\to 0,x\ne 0}\frac{f'(x)-f'(0)}{x-0}$$