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Set $$ f(x) = \begin{cases} x^2 \cdot \sin(1/x), &\text{when $x\neq 0$;}\\ 0, &\text{when $x=0$}. \end{cases}$$

Now we have to check whether $f''(x)$ is continuous at $x= 0$ and $''(0)$ exists or not. All I've done is calculating the $f''(x)$ as I don't know how to proceed. If you can help me about how to think this. I know that we can check RHS = value at point = LHS, but I cannot apply it here because of the sin function. I don't get it, thank you for helping.

P.s. what is reputation? Why do I need it to upload picture? :-o

max_zorn
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Rio
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  • The typical calculus definition of continuous at $a$ is that $f$ is continuous at $a$ if and only if $\lim_{x\to a}f(x)=f(a).$ Don't forget that this definition says three non-trivial things: 1. The limit exists. 2. The function is defined at $a$. 3. The two are equal. Any of these could fail, though they're no completely independent. – Adrian Keister Jul 10 '18 at 17:47
  • You might need the squeeze theorem to evaluate the limit. – Adrian Keister Jul 10 '18 at 17:47

1 Answers1

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We have $$ f'(x)=\begin{cases} 2\,x\sin\dfrac1x-\cos\dfrac1x&\text{if }x\ne0,\\ 0&\text{if } x=0. \end{cases} $$$f'$ is discontinuous at $x=0$, since $\lim_{x\to0}f'(x)$ does not exist. This implies that $f''(0)$ does not exist.

Julián Aguirre
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