# A real analysis proof question (related to sin(1/x))

The problem statement is

Let $$f(x) = \left\{\begin{array}{cc} x^4 \left(2 + \sin \frac 1 x\right) & x \ne 0 \\ 0 & x = 0 \end{array}\right.$$

(a)Prove that $f$ is differentiable on $\mathbb{R}$

(b)Prove that $f$ has an absolute minimum at $x=0$

(c) Prove that $f'$ takes both positive and negative values in every neighborhood of $0$.

This first two parts of the problem are pretty straightforward. The only problem I encountered was in the last part. I was not sure how to prove it. I know both $\sin(1/x)$ and $\cos(1/x)$oscillate near zero. For any interval around zero they are gonna take positive and negative values since they oscillates. But this was not consider as a "proof". Is there a way to do that more rigorously? Can I use intermediate value property?

BTW, $$f'(x) = 4x^3\left(2+\sin \frac 1 x\right)+x^2\cos \left(\frac 1 x\right)$$ when $x\neq 0$.

• Note that $\cos (1/x) = (-1)^n$ along the sequence $1/(n\pi).$ – zhw. Apr 14 '15 at 6:06
• Also note that for $|x|<1/12$ we have $$|12x^3|<x^2$$ which might be helpful. The term $4x^3(2+\sin(1/x))$ oscillates between $4x^3$ and $12x^3$ whereas the other term $x^2\cos(1/x)$ oscillates between $\pm x^2$. What can we conclude from this? – String Apr 14 '15 at 6:18

Since away from $0$ we have

$$f(x) = x^4\left(2 + \sin \frac 1 x\right)$$

we have

\begin{align*} f'(x) &= 4x^3 \left(2 + \sin \frac 1 x\right) + x^4 \left(\cos \frac 1 x\right) \left(-\frac 1 {x^2}\right) \\ &= 8x^3 + 4x^3 \sin \frac 1 x - x^2 \cos \frac 1 x \\ &= x^2 \left(8x + 4x \sin \frac 1 x - \cos \frac 1 x\right) \end{align*} (Note there's a sign error in the original post, but it doesn't actually matter). Now notice that the terms with $x$ tend to zero at the origin while the cosine term does not: The cosine term is dominant here. To be explicit, choose $x = 1 / (n \pi)$ with $n > 8$. The first two terms can be controlled as

$$\left|8x + 4x \sin \frac 1 x\right| \le 12 |x| < \frac 1 2$$

On the other hand, $\cos 1/x = \cos (n\pi) = \pm 1$ according to the parity of $n$. Hence, the derivative can be estimated either as

$$f'(x) > \frac 1 {(n\pi)^2} \cdot \left(1 - \frac 1 2\right) = \frac 1 {2 (n \pi)^2}$$ if $n$ is odd or

$$f'(x) < \frac 1 {(n \pi)^2} \cdot \left(-1 + \frac 1 2\right) = - \frac 1 {2(n\pi)^2}$$ if $n$ is even.