Epsilon-Delta-Definition: How to prove $f(x)=\begin{cases}x\cdot\sin\left(\frac{1}{x}\right) , x\neq 0 \\ 0, x=0\end{cases}$ is continuous in $x_0=0$

My attended proof:

In order to be continuous in $x_0$, one has to show that: $$\forall \epsilon >0 \exists \delta >0 \forall x\in \mathbb{R}: |x-x_0|<\delta \implies |f(x)-f(x_0)|<\varepsilon$$

I figured that $|f(x)-f(x_0)|=\big | x\cdot\sin\left(\frac{1}{x}\right) \big |\leq |x|<\varepsilon$. Furthermore $|x-x_0|=|x|<\delta$ How would I chose $\delta$? Couldn't I use $\delta = \varepsilon$?

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    $\begingroup$ Exactly.... So write it in the right order: let $\epsilon >0$. Pick $\delta=\epsilon$. Then, for all $x$ in $\mathbb R$ such that .... $\endgroup$ – N. S. Jul 20 '19 at 17:04

It would be clearer if you wrote $0$ everywhere you wrote $x_0$. But yes, you can choose $\delta = \varepsilon$. In this case, writing everything in the correct order of quantifiers, you would say: let $\varepsilon > 0$ be arbitrary, and choose $\delta = \varepsilon$. Let $x \in \Bbb{R}$ be such that $0< |x-0|< \delta = \varepsilon$. Then \begin{align} |f(x) -f(0)| &= \left|x\sin\left(\frac{1}{x} \right) - 0 \right|\\ & \leq |x| \\ &< \delta \\ &= \varepsilon. \end{align} Since $\varepsilon > 0$ was arbitrary, this shows $f$ is continuous at $0$.

  • $\begingroup$ Okay, cool! I could basically choose any $\delta$ right? $\endgroup$ – ParabolicAlcoholic Jul 20 '19 at 17:10
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    $\begingroup$ @ParabolicAlcoholic No! You're first given an arbitrary $\varepsilon > 0$; from now on, $\varepsilon$ is fixed. After this, you have to choose any $\delta$ satisfying $0 < \delta \leq \varepsilon$. So, for example if I give you $\varepsilon = 1$, you can't choose $\delta = 3$ for this proof; you need to choose $0< \delta \leq 1.$ $\endgroup$ – peek-a-boo Jul 20 '19 at 17:13

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