Show that $f$ is not increasing on any interval containing $0$ $f:R\to R$, $f(x)=x^2\sin(1/x)+x$ if $x\ne 0$ and $0$ if $x=0$
In the first part of this problem, I showed that $f'(0)>0$
The second part of the problem is this: Show that $f$ is not increasing on any interval containing $0$
I tried to find one positive number $h$ arbitrarily close to $0$, and one negative number $k$ arbitrarily close to $0$, such that $f(h)$ & $f(k)$ were both $\lt 0$ but I could not figure out how to do so. Here is my attempt:
Let $k\to 0^-$. As long as $|k|\lt 1$, we have $|k^2|\lt |k|$ so $f(k)\lt 0$. 
Let $h\to 0^+$. Note $1/h\to \infty$, so $\sin(1/h)$ cycles between $-1$ and $1$. To find the minimum of $f(h)$, choose $h\gt 0$ s.t. $\sin(1/h)=-1$. $|h|\lt1$ so $h^2<h \implies 0\lt h-h^2=f(h)$.
Clearly this does not prove what I want. How should I approach this problem?
 A: Clearly we have $f'(x) = 2x\sin (1/x) + 1 - \cos (1/x)$ and let's put $y = 1/x$ so that $f'(x) = (2/y)\sin y - \cos y + 1 = g(y) + 1$. Now as $x \to 0^{+}$ we have $y \to \infty$. We will show that as $x \to 0^{+}$ there are infinitely many values of $x$ for which $f'(x) < 0$. This involves that as $y \to \infty$ there are infinitely many values of $y$ for which $g(y) < -1$.
We need to analyze the maxima/minima of $g(y)$. Now $g'(y) = (2/y)\cos y - (2/y^{2})\sin y + \sin y$ and hence $g'(y) = 0$ implies that $$\cos y = (\sin y)\left(\frac{2 - y^{2}}{2y}\right)\tag{1}$$ and then at these points of extrema we have $$g(y) = \frac{2\sin y}{y} - \cos y = \frac{2\sin y}{y} - (\sin y)\left(\frac{2 - y^{2}}{2y}\right) = (\sin y)\left(\frac{2 + y^{2}}{2y}\right)\tag{2}$$ Note that from graphical considerations it is easy to show that equation $(1)$ has infinite many solutions and the values $y$ satisfying this equation increase without bound. In other words there are many large values of $y$ (exceeding any given bound) which satisfy equation $(1)$.
Next from $(1)$ we can see that $$\sin^{2}y = \frac{1}{1 + \cot^{2}y} = \dfrac{1}{1 + \left(\dfrac{2 - y^{2}}{2y}\right)^{2}} = \frac{4y^{2}}{y^{4} + 4}$$ and then from $(2)$ we get $$g(y) = \pm\frac{2 + y^{2}}{\sqrt{y^{4} + 4}}$$ where the sign $\pm$ is chosen same as the sign of $\sin y$. These values of $g(y)$ are its extrema. Note that in absolute value the expression for $g(y)$ is always greater than $1$ if $y > 0$. Hence it follows that there will be infinitely many values of $y$ corresponding to solutions of equation $(1)$ for which $\sin y$ will be negative so that $g(y) < -1$.
We have thus established that $f'(x) < 0$ for infinitely many values of $x$ as $x \to 0^{+}$. Now it is also easily seen that as $x \to 0^{+}$ the term $2x\sin (1/x) \to 0$ and $\cos (1/x)$ oscillates between $-1$ and $1$ so that the expression for $f'(x)$ oscillates between $0$ and $2$ and hence there are infinitely many values of $x$ for which $f'(x) > 0$.
Thus in any neighborhood of $0$ we have $f'(x) > 0$ for some values of $x$ and $f'(x) < 0$ for some other values of $x$. It follows that $f(x)$ is not increasing in any neighborhood of $0$.
Note: The above solution is inspired by an exercise problem (hints were provided in the problem) from "Calculus" by Michael Spivak.
A: You won't find such $k$ and $h$, because the following can be proved.  Let $f$ be defined on an open interval $J$ containing $0$ and let $f$ be differentiable in $0$ and $f'(x)>0$.  Then there exists $\epsilon>0$ such that for all $x\in]-\epsilon,\epsilon[\cap J$ we have
$$x<0\Rightarrow f(x)<0\quad\text{and}\quad x>0\Rightarrow f(x)>0.$$
So let both $h$ and $k$ be positive.
