# Awkward behavior of $x^2 \sin\frac{1}{x}$ at $x=0$?

According to p. 107 of the book Advanced Calculus by Fitzpatrick,

Assuming that the periodicity and differentiability properties of the sine function are familiar, the following is an example of a differentiable function having a positive derivative at $x = 0$ but such that there is no neighborhood of $0$ on which it is monotonically increasing: $$f(x) = \begin{cases}x^2\sin 1/x & \text{if}\ x\neq 0 \\ 0 & \text{if}\ x = 0\end{cases}$$ The source of this counterintuitive behavior is that the derivative $f'$ is not continuous at $x = 0$.

There are two confusing things about it:

1. a. By the definition of derivative (at $x=0$), $$\lim_{x\to 0} \dfrac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \dfrac{x^2 \sin \bigl(\frac{1}{x}\bigr)}{x} = 0,$$ since $\bigl\lvert\sin\bigl(\frac{1}{x}\bigr)\bigr\rvert \le 1$.

b. By use of rules for the derivative of products and quotients, $$f'(0) = \biggl[ 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr) \biggr]_{x=0}$$ which is not defined since $\cos\bigl(\frac{1}{0}\bigr)$ is not defined.

So why the text says that its derivative exists and its value is $>0$?

2. The function $f(x)$ is monotonically increasing because it is an odd function so for any neighborhood abound $x=0$, $f(x_1>0)>f(x_2<0)$. So why the text says otherwise?

• The statement about $f'(0)>0$ is of course false, as you've seen yourself. Presumably the author meant something like $x/2+ x^2 \sin(1/x)$ instead. Then $f'(0)=1/2$, but in any neighbourhood of $x=0$, the derivative $f'(x)$ assumes both positive and negative values. Sep 16, 2016 at 12:51
• @HansLundmark, I have been uploaded the image of text before David's edit. It's the first time I see an error in this amazing book!
– user231343
Sep 16, 2016 at 13:22
• Jan 14, 2017 at 14:29

1.- (a) is correct so $f'(0)=0$, but your conclusion on 1.- (b) is wrong. Notice that you are trying to compute the derivative of $f$ at $0$ by using the formula for $f$ on points different from zero. The formula $$f'(x) = 2x \sin \biggl(\frac{1}{x}\biggr) + x^2 \biggl(-\frac{1}{x^2} \cos \biggl(\frac{1}{x}\biggr)\biggr)$$ holds for all $x\neq0$ but, as you know, this formula is undefined at $0$. Therefore, you can conlude that the derivative of $f$ is not continuous at $0$, since you cannot approximate $f'(0)$ by $f'(x)$ for $x$ near $0$. In other words, $$f'(0) \neq \lim_{x\rightarrow 0} f'(x).$$

2.- In general, odd functions are not monotonically increasing. For example, $$x^3 - x$$ is odd, but it is not monotonic. Indeed, $x^3-x$ is increasing on the intervals $(-\infty,-\tfrac{1}{\sqrt{3}})$ and $(\tfrac{1}{\sqrt{3}},\infty)$, but is decreasing on $(-\tfrac{1}{\sqrt{3}},\tfrac{1}{\sqrt{3}})$.

Back to $f(x) = x^{2}\sin(\tfrac{1}{x})$, you can tell that it is not increasing because it is positive on the intervals $(\tfrac{1}{2k\pi},\tfrac{1}{(2k-1)\pi})$ but negative on $(\tfrac{1}{(2k+1)\pi},\tfrac{1}{2k\pi})$ for every $k\neq0$. Since every neighborhood of $0$ contains infinitely many intervals of those types, we conclude that $f(x)$ has infinitely many changes of sign on any neighborhood of $0$. Thus, it is not monotonic near $0$.