1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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 $x=0$. $\endgroup$
    – hardmath
    Dec 15, 2018 at 4:06

2 Answers 2

5
$\begingroup$

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}

$\endgroup$
3
$\begingroup$

hint

$$f''(0)=\lim_{x\to 0,x\ne 0}\frac{f'(x)-f'(0)}{x-0}$$

$\endgroup$
1
  • $\begingroup$ @hardmath Done. thank you. $\endgroup$ Dec 14, 2018 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.