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Is the function $y=x/2 + x^{2}\sin(1/x)$ monotonic near $0$?

The derivative $f'$ obviously goes positive and negative near $0$, because $$f'(x)= \frac12 + 2x\sin(1/x) - \cos(1/x))$$ Does that mean that $f$ is not monotonic near $0$?

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    If $f'$ takes both positive and negative values in every neighborhood of $0$, then $f$ cannot be monotonic in any neighborhood of $0$.2012-12-18
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    Think of it this way: if the derivative takes on both positive and negative values near 0, then the function both increases and decreases near 0. So, it isn't either just increasing or just decreasing no matter how close to 0 we get, meaning it can't be monotone at 0.2012-12-18
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    Thank you guys... i wasn't sure about f but for small x 2xsin in f ' is ~0 so 1/2 - cos(1/x) i think can be both + and - (depending on cos)...2012-12-18
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    Nevertheless, you of course do have $f(x)>f(0)$ for $x>0$ and $f(x)2012-12-18
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    Show (by continuity) that for any $\epsilon$, there is an $a$ with $|a|\lt \epsilon$, and an *interval* about $a$, such that $f'(x)\lt 0$ in that interval. So by Mean Value Theorem, $f$ is decreaing in that interval. Repeat with $f'(x)\gt 0$.2012-12-18

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It may be appropriate to now summarize what is spread across comments into an answer.

The function $$f(x) = \begin{cases}\tfrac x2+x^2\sin(1/x)&\text{if }x\ne 0\\ 0&\text{if }x=0\end{cases}$$ has one property that may make it look like being monotonic near $0$: For $0<|x|<\frac 12$ we have $|x^2\sin(1/x)|\le|x^2|<\left|\frac x2\right|$, hence $f(x)>0$ for $0at $0$, but I'm not sure if that concept is in use). However, we say that $f$ is monotonic near $0$ if there exists an open neighbourhood $U$ of $0$ such that the restriction $f|_U$ of the given function $f$ to that neighbourhood $U$ is monotonic. As you already noted, for $x\ne0$ we have $$f'(x)= \frac 12 + 2x\sin(1/x) - \cos(1/x).$$ Let $U$ be an open neighbourhood of $0$. For $n\in \mathbb N$ big enough, we have $\xi:=\frac1{2n\pi}\in U$. Then we check that $f'(\xi)=-\frac12<0$. Since $U$ is open, we have $\xi+h\in U$ for all sufficiently small positive $h$. By the definition of derivative, for all sufficiently small positive $h$ we have that $\left|\frac{f(\xi+h)-f(\xi)}{h}-f'(\xi)\right|<\frac12$ and hence $f(\xi+h)0$ because $0<\xi<\frac12$. Then from $f(0)f(\xi+h)$ we see that $f|_U$ is not monotonic.