Prove that $\lim_{x \to 0} x \sin\left(\frac1x\right)=0$

Question

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..

Answer 1

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.

Answer 2

You don't really need to use the $\varepsilon - \delta$ definition, unless you are explicitly told to: $$ 0 \leq \left| x \sin\left( \frac 1x\right) - 0\right|\leq |x| \Rightarrow \lim_{x\to 0} x \sin\left( \frac 1x\right) = \lim_{x\to 0}|x| = 0. $$

Answer 3

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.