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say I have the function $$ f(x)=\begin{cases} x^{3}\sin(\frac{5}{x}) & x\ne0\\ 0 & x=0 \end{cases} $$ I want to prove it is differentiable at $0.$ I first show that it is continuous by: $$ \lim_{x\to0}f(x)=x^{3}\sin(\frac{5}{x})=0 $$ That is because I have a bounded function multiplied by 0. Now i need to show if f is differentiable and if so,is $f'$ continuous? I started by doing $$ f'(0)=\lim_{h\to0}\frac{(0+h)^{3}\sin(\frac{5}{0+h})-(0)^{3}\sin(\frac{5}{0})}{h} $$ But I can't seem to understand how do I go about the $\sin (\frac{5}{0})$ part because that doesn't exist. Do i take 2 limits of $ x\to 0 $ and $h\to 0 $ in that case? As for the second part of showing that f' is continuous I just differentiated with the product rule and got that: $$ f'(x)=3x^2\sin(\frac{5}{x})-5x\cos(\frac{5}{x}) $$ And after checking both sides I saw that one of them is negative and the other is positive, hence it has different limits at the two sides. How do I solve these problems by definition?

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    $\begingroup$ Why did you write $0^3\sin\frac 50$? The hypothesis does not say that $f(0)=0^3\sin\frac 50$. $\endgroup$
    – user239203
    Commented Jun 22, 2021 at 15:30
  • $\begingroup$ So instead I just say that $$ f'(0)=\lim_{h\to0}\frac{(0+h)^{3}\sin(\frac{5}{0+h})-0}{h}=\frac{h^{3}sin(\frac{5}{h})}{h}=h^{2}\sin(\frac{5}{h})=0$$ $\endgroup$ Commented Jun 22, 2021 at 15:39
  • $\begingroup$ @DannyBlozrov Yes, that is the correct way to evaluate $f'(0)$. $\endgroup$
    – Mark Viola
    Commented Jun 22, 2021 at 15:49

4 Answers 4

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The defintion of the derivative at $0$ is :$$f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h} $$ You are given that $f(0)=0$ from here the limit is easy to evaluate.
For other $x$ you can use the product and chain rule .

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    $\begingroup$ (+1) for a complete solution that actually addresses the OP's first question. $\endgroup$
    – Mark Viola
    Commented Jun 22, 2021 at 15:48
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You have that \begin{eqnarray*} f: I\subset \mathbb{R} &\to& \mathbb{R}\\ x&\mapsto& f(x)=\begin{cases}x^{3}\sin\left(\frac{5}{x}\right), &\quad x\not=0\\ 0, &\quad x=0 \end{cases} \end{eqnarray*} If you want to prove that $f$ is differentiable at $0$, you do not need to start by proving that $f$ is continuous at $0$. Of course, if $f$ is not continuous at $0$, then $f$ is not differentiable at $0$. But, it is not what is requested in the problem.

You need to prove that $$\lim_{h\to 0} \frac{f(0+h)-f(0)}{h}$$there exists and if that limit there exists, then $$f'(0)=\lim_{h\to 0} \frac{f(0+h)-f(0)}{h}.$$

Now, you know that if $x=0$, so $f(0)=0$ and for all $x\not=0$, you have that $f(x)=x^{3}\sin\left(\frac{5}{x}\right)$.

Hence, $$\lim_{h\to 0}\frac{h^{3}\sin\left(\frac{5}{h}\right)-0}{h}=\lim_{h\to 0}h^{2}\sin\left(\frac{5}{h}\right)=0. \quad (1)$$

Note that $(1)$ it's clear, because $$-1\leqslant \sin\left(\frac{5}{h}\right)\leqslant 1$$ $$-h^{2}\leqslant h^{2}\sin\left(\frac{5}{h}\right)\leqslant h^{2}$$ $$\lim_{h\to 0} -h^{2}\leqslant \lim_{h\to 0} h^{2}\sin\left(\frac{5}{h}\right)\leqslant \lim_{h\to 0} h^{2}$$ $$0 \leqslant h^{2}\sin\left(\frac{5}{h}\right)\leqslant 0$$ Therefore, $$\lim_{h\to 0} h^{2}\sin\left(\frac{5}{h}\right)=0$$ and $f$ is differentiable at $x=0$ with $f'(0)=0$.

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  • $\begingroup$ (+1) for a complete solution that actually addresses the OP's first question. $\endgroup$
    – Mark Viola
    Commented Jun 22, 2021 at 15:48
  • $\begingroup$ Thank you very much,understood it all. As for showing that f' is continuous I can just use the product and chain rule to find $f'(x) for any $ x\ne 0 $ and I know that $ f'(0)=0 $ ,now I just have to compare the limit to 0 and see if its the same? $\endgroup$ Commented Jun 22, 2021 at 16:01
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To check differentiability at $0$ you should essentially calculate the limit $$ \lim_{h \to 0} \frac{f(h) -f(0)}{h} = \lim_{h\to 0} \frac{h^3 \sin(5/h)}{h} = \lim_{h \to 0} h^2 \sin \left( \frac{5}{h} \right)$$.

Observe that $$0 \le \left\vert h^2 \sin \left( \frac{5}{h} \right) \right\vert \le h^2$$ So by Squeeze_theorem the limit exists and equal to $0$.

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  • $\begingroup$ Not really following why, that's a must have condition that checks if it is continuous, why does it mean its differentiable? $\endgroup$ Commented Jun 22, 2021 at 15:44
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    $\begingroup$ This fails to address the main question. $\endgroup$
    – Mark Viola
    Commented Jun 22, 2021 at 15:47
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When you look at the product rule, you need that both factors are differentiable. But $\sin(5/x)$ is not differentiable in $0$, so you cannot apply this rule to your example.

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