Show $f(x) = x^2\sin{\frac{1}{x}} + \frac{x}{2}$ is not increasing on any open interval containing $0$ $f(x) = x^2\sin{\frac{1}{x}} + \frac{x}{2}$ for $x\not= 0$ and $f(0) = 0$.
Show $f$ is not increasing on any open interval containing $0$.
At first glance, we notice $f'(x) \le 0$ for some $x \in I$ where $I$ is any open interval. We can also recognize $f'(0) > 0$ by directly plugging in $0$ into the definition of a limit:


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*$f'(x) = \lim_{x\to x_0}{\frac{f(x)-f(x_0)}{x-x_0}}$


We know that if $f'(0)>0$, then $f'(x)<0$ either above or below $0$.
However, at this point I get stuck on how to continue.
 A: Clearly you can note that $f'(0) = 1/2$ and $$f'(x) = 2x\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) + \frac{1}{2}$$ if $x \neq 0$. As $x \to 0$ the first term $2x\sin(1/x)$ tends to $0$ and $\cos(1/x)$ oscillates between $-1$ and $1$. So as $x \to 0$ the derivative $f'(x)$ oscillates between around $-1 + 1/2$ and $1 + 1/2$ i.e. between around $-1/2$ and $3/2$. In more technical symbolism we write $$\liminf_{x \to 0}f'(x) = -\frac{1}{2},\,\limsup_{x \to 0}f'(x) = \frac{3}{2}$$ So any neighborhood of $0$ contains some points $a$ where $f'(a) > 0$ and some other points $b$ where $f'(b) < 0$. Thus it is not possible that $f$ in increasing in any neighborhood of $0$.
Note that $f'(0) = 1/2 > 0$ and yet there is no interval around $0$ in which $f$ is increasing. To quote Hardy in this matter:
"If $f'(x)$ is positive at a single point $x_{0}$, then we can prove that $f(x_{1}) < f(x_{0}) < f(x_{2})$ for points $x_{1}, x_{2}$ in a certain neighborhood of $x_{0}$ with $x_{1} < x_{0} < x_{2}$. But this does not prove that there is any interval including $x_{0}$ throughout which $f(x)$ is a steadily increasing function, for the assumption that $x_{1}$ and $x_{2}$ lie on opposite sides of $x_{0}$ is essential to our conclusion."
However if $f'(x) > 0$ for all points $x$ of an interval $I$ then $f(x)$ is strictly increasing on interval $I$.
