Saturday 27 September 2014

continuity - Checking whether a function is continuous or not



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


Answer



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.


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