Prove that $x\left(1+2x\sin\left(\frac{1}{x}\right)\right)$ is differentiable

Question

I am having difficulties proving that

$f:\mathbb{R}\to \mathbb{R}$, $x\mapsto\begin{cases}x\left(1+2x\sin\left(\frac{1}{x}\right)\right),&\text{if }x\neq0\\0,&\text{if }x=0\end{cases}$ is differentiable with $f'(0) > 0$.

Also show that $f$ is not strictly monotonically increasing in neighbourhood of point $0$.

So I must show that $(-\varepsilon, \varepsilon)$ for every $\varepsilon > 0$ has an interval in which $f$ is strictly monotonically decreasing and can use the properties that $\sin'(x) = \cos(x)$, $\cos'(x) = -\sin(x)$ and $\sin(2\pi n) = 0$, $\cos(2 \pi n) = 1$.

I can't seem to figure it out. Any help would be highly appreciated.

Answer 1

Well, for every $x\neq0$ $f$ is obviously differentiable since it is product, sum and synthesis of function that are differentiable. We have to give a closer look to $x_0=0$.

By definition, we have: $$f'(0)=\lim_{x\to x_0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x\left(1+2x\sin(\frac{1}{x})\right)}{x}=\lim_{x\to0}\left[1+2x\sin\left(\frac{1}{x}\right) \right]$$

So, $f'(0)>0$ if and only if the above limit does exist and is strictly positive. For the calculation of the above limit, we have:

$$\left|x\sin\left(\frac{1}{x}\right)\right|\leq|x|\cdot1\mbox{, since $\sin\phi\leq1$}$$ So, we have: $$0\leq\lim_{x\to0}\left|x\sin\left(\frac{1}{x}\right)\right|\leq\lim_{x\to0}|x|=0\Rightarrow\lim_{x\to0}\left|x\sin\left(\frac{1}{x}\right)\right|=0\Rightarrow\lim_{x\to0}x\sin\left(\frac{1}{x}\right)=0$$ So, for our first limit, we have: $$\lim_{x\to0}\left[1+2x\sin\left(\frac{1}{x}\right) \right]=1+2\lim_{x\to0}x\sin\left(\frac{1}{x}\right)=1+2\cdot0=1$$

So: $$f'(0)=1>0$$ which is what has been requested.

EDIT:(I did not see the last part of the question) For every other $x\neq0$, we have: $$f'(x)=1+2x\sin\left(\frac{1}{x}\right)+x\left(2\sin\left(\frac{1}{x}\right)-\frac{2}{x}\cos\left(\frac{1}{x}\right)\right)$$ or, shorter: $$f'(x)=1+4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)$$

So, let $\epsilon>0$ and let $x_k=\frac{1}{k\pi}$. There is, by Archimedes-Eudoxus Principle, a $k_0\in\mathbb{N}$, such that: $$\frac{1}{k\pi}<\epsilon$$ for every $k\geq k_0$. So, for every $k>k_0$, we have: $$f'(x_k)=1+\frac{4}{k\pi}\sin(k\pi)-2\cos(k\pi)=1+2\cdot(-1)^{k}=\left\{\begin{array}{ll}3&\mbox{if $k$ is even}\\-1&\mbox{if $k$ is odd}\end{array}\right.$$

So, in any open neighbourhood $(-\epsilon,\epsilon)$ of $0$, $f'$ takes both positive and negative values and, hence, $f$ is not strictly monotonous there.

Answer 2

Hints:

1. Compute $f'(x)$ for $x \ne 0$.

2. $\frac{f(x)-f(0)}{x-0}=1+2x \sin(1/x) \to 1$ for $x \to 0$.

3. $f'(\frac{1}{n \pi})=(-1)^{n+1}$ for all $n \in \mathbb N$