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I am looking for a hint or feedback on what I've already done, not a full solution. $f=t\sin{\left(\frac{1}{t}\right)}$ for $t\ne 0$, $f(0)=0$,
My idea is that I only have to worry about the steep parts.

My approach:
Proof sketch: Let $\epsilon > 0 $ be arbitrary, we seek to find $\delta$ such that $\forall x,y \in [0,1]$, $|x-y|<\delta \Longrightarrow |x\sin{\left(\frac{1}{x}\right)}- y\sin{\left(\frac{1}{y}\right)}|<\epsilon$

\begin{align*} & t\sin{\left(\frac{1}{t}\right)} = \frac{2}{\pi(4k+1)} \text{ is true for } t =\frac{2}{\pi(4k+1)} \\ & t\sin{\left(\frac{1}{t}\right)} = - \frac{2}{\pi(4k+3)} \text{ is true for } t =\frac{2}{\pi(4k+3)} \end{align*} and \begin{align*} &\text{distance}\left(f_1\left(\frac{2}{\pi(4k+1)}\right), f_1\left(\frac{2}{\pi(4k+3)}\right)\right) = \\ & \frac{2}{\pi(4k+1)} - \left(- \frac{2}{\pi(4k+3)} \right) =\\ & \frac{16k+8}{\pi (4k+1)(4k+3)}=\\ & \left(\frac{4}{\pi}\right)\frac{4k+2}{(4k+1)(4k+3)} & \end{align*} Define $\epsilon(k) = $ such $k$ so that $$ \left(\frac{4}{\pi}\right)\frac{4k+2}{(4k+1)(4k+3)} < \epsilon $$

So as long as $\delta = \frac{2}{\pi(4\epsilon(k)+1)} - \frac{2}{\pi(4\epsilon(k)+3)}= \frac{4}{\pi(4\epsilon(k)+1)(4\epsilon(k)+3)}$ we should be good? I am trying to focus on the pre image of the steepest parts of the function and then looking at the width of that. I am pretty brain dead right now... I feel like this delta works, but I need more/better justification...

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Rustyn, hi! You only need to worry about what happens at $0$; I assume you are defining $f(0)=0$, right? And you know that sine is bounded. – Andrés E. Caicedo Aug 28 '13 at 6:18
@AndresCaicedo Yes, that's right f(0) = 0, and I played around with the idea that sine is bounded, that gives that every delta works for epsilon greater than or equal to 2. – Rustyn Aug 28 '13 at 6:20
@AndresCaicedo I think I will assume that I can use the theorems from analysis I. I will have to prove them first, grrrrr – Rustyn Aug 28 '13 at 6:31
@StefanSmith thanks, but I've already finished the problem – Rustyn Aug 29 '13 at 16:41
up vote 3 down vote accepted

For the part $2$ of @copper.hat's, you might use this point that if $0<|x|<\delta$ and get $\epsilon=\delta$ then $$\left|x\sin(1/x)\right|=|x||\sin(1/x)|\leq|x|<\epsilon$$

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$+1 \ddot \smile$ – amWhy Aug 29 '13 at 0:02

I'm not sure what you are trying to do above.

Try splitting the proof into two parts:

Part 1: take $t \in (0,1]$. This is straightforward as $t \mapsto t$, $t \mapsto \frac{1}{t}$, and $\sin$ are continuous, and so their compositions and products are too.

Part 2: Determine continuity at $t=0$. Presumably you have $f(0) = 0$, so you need to show that for any $\epsilon>0$, you can find a $\delta>0$ such that if $t \in [0,\delta)$ then $|f(0)-f(t)| < \epsilon$. This should be straightforward too.

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It seems Rustyn is trying to argue that the function is uniformly continuous. But this follows automatically from mere continuity, by compactness. – Andrés E. Caicedo Aug 28 '13 at 6:21
I could already have done part 1 and part 2, if I had proved those facts that you are using in part 1 – Rustyn Aug 28 '13 at 6:24
@copper.hat grrrrrr Your answer doesn't work if the student is taking analysis I and analysis II at the same time. LOL!!!! – Rustyn Aug 28 '13 at 6:27
Proving Part 1 from first principles seems pretty messy? – copper.hat Aug 28 '13 at 6:39
@RustynYazdanpour: But his points are enough for an student who gets Calculus I to find out the problem. – Babak S. Aug 28 '13 at 6:39

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