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Hi everyone im struggling a bit with this question. I kinda get the idea of epsilon-delta proof, and have seen you can choose $\epsilon = \delta$ I don't get why tho. Could someone explain that to me?

Another way I thought of the problem, was to show that if there $\exists$ a relation between $\epsilon$ and $\delta$ the limit must exist.

My way of tackling the problem:

If we start with $|f(x)-L|< \epsilon \implies |(x\cdot sin(\frac{1}{x}))-0|<\epsilon$

Now we want the expression on the form as $|x-0|=|x|<\delta$

$$|x\cdot sin\left(\frac{1}{x}\right)-0| <\epsilon$$ $$=|x|\cdot|sin\left(\frac{1}{x}\right)|<\epsilon$$ $$ =|x| < \frac{\epsilon}{sin(\frac{1}{x})}$$ Thus $$\delta =\frac{\epsilon}{sin(\frac{1}{x})}$$

Since there is a relation between $\epsilon$ and $\delta$ the limit must exist. Dont know if it is correct way of proving it or not..

Rodrigo de Azevedo
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Kevz123
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  • The idea is that the first factor $x$ tends to zero and the second factor $\sin(1/x)$ oscillates wildly but stays in the range $[-1,1]$. So you have something almost zero times something bounded. – Daron Feb 22 '22 at 13:29
  • You cannot "start with" $|f(x)-L|< \epsilon \implies |(x\cdot sin(\frac{1}{x}))-0|<\epsilon$ because that is what you are trying to prove. You start with what you know and deduce what you want to know. – Daron Feb 22 '22 at 13:30
  • I edited the title. Please edit the body. – Rodrigo de Azevedo Feb 22 '22 at 13:31
  • Just saying, are you only in the reals? Because the answer depends on how you approach $0$ in the complex plane. – Diger Feb 22 '22 at 13:33
  • So I guess im just terrible wrong haha..

    It's only in the reals @Diger.

     – Kevz123 Feb 22 '22 at 13:34

3 Answers3

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That makes no sense. You are supposed to prove that, for every $\varepsilon>0$, there is some $\delta>0$ depending only upon $\varepsilon$ such that$$0<|x|<\delta\implies\left|x\sin\left(\frac 1x\right)\right|<\varepsilon.$$Your $\delta$ does not depend only upon $\varepsilon$.

On the other hand, $\delta=\varepsilon$ will work.

José Carlos Santos
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Do note that $\delta$ must depend entirely on $\varepsilon$ whereas yours depends also on $x$ through $\sin\left(\frac{1}{x}\right)$. Therefore it doesn't work.

Sometimes when dealing with limits by definition it is good to see why you did it a certain way with rules beforehand (not at all always useful, but sometimes it is, like in this case). Notice that for $$\lim_{x\to0}\; x\sin\left(\frac{1}{x}\right)$$ You used to say bounded times $0$ is equal to $0$. In this case you can use that statement, to notice that the bounded term is $|\sin\left(\frac{1}{x}\right)|\leq 1$ and so $$\left|x\sin\left(\frac{1}{x}\right)\right|\leq |x|$$ So if $|x|<\varepsilon$, we have what we were looking for. And we see that $\delta=\varepsilon$ works.

ALakatos
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  • how do I know that if $|x| < \epsilon$ is it just an assumption? – Kevz123 Feb 22 '22 at 13:43
  • Indeed, let $\varepsilon>0$ and you need to prove there is a $\delta$ so that if $0<|x|<\delta$ then $|x\sin(\frac{1}{x})|<\varepsilon$. You can choose $\delta=\varepsilon$ and see that you have effectively bounded what you wanted. – ALakatos Feb 22 '22 at 13:51
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You don't really need to use the $\varepsilon - \delta$ definition, unless you are explicitly told to: $$ 0 \leq \left| x \sin\left( \frac 1x\right) - 0\right|\leq |x| \Rightarrow \lim_{x\to 0} x \sin\left( \frac 1x\right) = \lim_{x\to 0}|x| = 0. $$

PierreCarre
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