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Let $f: \mathbb R \rightarrow \mathbb R$ defined by $$f(x) := \begin{cases}x^2 \sin \frac 1 x\ & x \neq 0\\ 0\ & x = 0\end{cases}$$

Show $f$ is differentiable on $\mathbb R$:

Let $\epsilon > 0$.

$x_0 \neq 0$: $\left\lvert\frac {x^2 \sin \frac 1 x - x_0^2 \sin \frac 1 {x_0}}{x-x_0} \right\rvert \le \epsilon$

I'm thinking that since the fraction is continuous there exists $\delta > 0$ satisfying the above for $x \in [-\delta,\delta] / \{0\}$ ?

$x_0 = 0$: $\left\rvert\frac {x^2 \sin \frac 1 x - 0} {x} \right\rvert \le \epsilon$

How can I evaluate this limit ? I cannot say the fraction is continuous at $0$ ?

Also I must find the derivative of $f$ and show it is not continuous at $x=0$, but I'm stuck.

Shuzheng
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  • Do you require an $\varepsilon$-$\delta$ proof? – Git Gud Apr 09 '14 at 09:21
  • No probably not, but it is the formal way of doing it ? – Shuzheng Apr 09 '14 at 09:24
  • It's hard to believe an $\varepsilon$-$\delta$ proof would be asked to prove this. It's probably safe to assume that you can use properties like $g,h\text{ differentiable}\implies gh\text{ differentiable}$ and so on. So you really only have to worry about proving differentiability at $0$ and to do this you just need to evaluate a limit. – Git Gud Apr 09 '14 at 09:28
  • What are the points, where you may have troubles/are not sure that function is differentiable? It is only $x=0$, obviously. – sas Apr 09 '14 at 09:28
  • Yes, but it is also this limit I'm having trouble evaluating. – Shuzheng Apr 09 '14 at 09:37
  • I added a proof that your $f$ is not differentiable on $\mathbb{R}$. Maybe you or the text where this function is taken from committed a typo. – Alessandro Flati Apr 09 '14 at 09:53

2 Answers2

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Proving differentiability at $0$ is equivalent to proving that $\lim \limits_{x\to 0}\left(\dfrac {f(x)-f(0)}{x-0}\right)$ exists and is finite.

This is turn is equivalent to proving the same hods for $\lim \limits_{x\to 0}\left(x\sin \left(\dfrac 1 x\right)\right)$.

The last statement above is equivalent to $$\forall \varepsilon >0\,\exists \delta >0\,\forall x\in D_f\left(0<|x-0|<\delta \implies \left|x\sin \left(\dfrac 1 x\right)\right|<\varepsilon\right).$$

To prove it recall that $\forall \alpha \in \mathbb R(|\sin(\alpha)|<1)$. Can you find $\delta$ given $\varepsilon$?

Git Gud
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    I suggest to rather use $|\sin\alpha|\le 1$ – Hagen von Eitzen Apr 09 '14 at 09:51
  • Ahh, I have $|x \sin(\frac 1 x)| \le |x|$. So I should choose $|x| \le \epsilon$ ?. I have calculated $f^{'}(x)$ as $2x \sin \frac 1 x - \cos \frac 1 x$ for $x \neq 0$ and $0$ for $x = 0$. Is there a clever way of showing that continuity doesn't hold at $x = 0$ ? – Shuzheng Apr 09 '14 at 09:53
  • @HagenvonEitzen Of course, much better. Thank you. – Git Gud Apr 09 '14 at 09:53
  • @user111854 Your question confuses me: "So I should choose $|x|≤ϵ$". You don't choose $\varepsilon$ and you don't choose $x$, what you choose is $\delta$. – Git Gud Apr 09 '14 at 09:55
  • My point is that given $\epsilon > 0$ I must choose $\delta > 0: 0 < |x| \le \delta$ satisfies $|x| \le \epsilon$. But in this case I can choose $\delta = \epsilon$ ? – Shuzheng Apr 09 '14 at 09:57
  • @user111854 That's correct. As for the continuity, find the derivative for $x\neq 0$ and check the limit at $0$. – Git Gud Apr 09 '14 at 10:00
  • $2x \sin \frac 1 x \le 2x \rightarrow 0$ for $x \rightarrow 0_+$ . But $\cos \frac 1 x$ will vary between $-1$ and $1$ for $x \rightarrow 0_+$. This is a proof ? – Shuzheng Apr 09 '14 at 10:04
  • Ahh, nice. I came up with one too. Let $\epsilon = \frac 1 2$. Then $|2x \sin \frac 1 x - \cos \frac 1 x| \le \epsilon \iff -\frac 1 2 + \cos \frac 1 x \le 2x \sin \frac 1 x \le \frac 1 2 + \cos \frac 1 x$. Now suppose $f^{'}$ is continuous at $x=0$ then there exists $\delta > 0 : 0 < |x| \le \delta$ satisfies the above. But then choose $x = \frac 1 {k \pi}$ for $k \in {1,3,5,...}$ to get $-\frac 3 2 \le 0 \le -\frac 1 2$. – Shuzheng Apr 09 '14 at 10:16
  • Thanks a lot @GitGud - You have been a big help. – Shuzheng Apr 09 '14 at 10:30
  • @user111854 You're welcome. – Git Gud Apr 09 '14 at 10:31
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Lemma: if $\;h,g\;$ are real functions defined in some neighborhood $\;D_0\;$ of $\;x_0\in\Bbb R\;$ s.t.

$$\begin{cases}\lim_{x\to x_0}h(x)=0\\{}\\\exists\,M\in \Bbb R\;\;s.t.\;\;\forall\,x\in D_0\;,\;\;|g(x)|\le M\end{cases}\;\;,\;\;\text{then}\;\;\lim_{x\to\ x_0}h(x)g(x)=0$$

The proof is boringly simple, and it gives you what you want since

$$\frac{f(x)-f(0)}{x-0}=x\sin\frac1x$$

and $\;\sin\frac1x\;$ is bounded in a neighborhood of zero....

DonAntonio
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