Prove or disprove that $f(x) = \sin x/x$ is uniformly continuous over the interval $(0,1)$? I am thinking of $\lim_{x \to 0} \ {f(x)} = 1$ but I am confused as $x=0$ is not in the domain and also I want to write an $\epsilon$-$\delta$ proof. So any help is much appreciated!
Thanks! 
 A: Ittay Weiss described the right way to solve this problem. If you really want an $\varepsilon$-$\delta$ solution, you can do the following.
Given $\varepsilon > 0$, let $\delta = \varepsilon$. Now for $x,y\in (0,1)$ with $|x-y| \leq \delta$, we have
$$|f(x) - f(y)| = |(x-y) f'(\xi)| = |x-y| \cdot |f'(\xi)| \leq \delta |f'(\xi)|,$$
where $\xi \in (0,1)$. Note that 
$$f'(x) = \frac{x\cos x - \sin x}{x^2}.$$
 This function takes negative values on $(0,1)$ since $\tan x > x$ (for $x\in (0,1)$). Also
$$\frac{x\cos x - \sin x}{x^2} \geq \frac{x\cos x - x}{x^2} = \frac{\cos x - 1}{x} \geq -1.$$
That is, $f'(\xi) \in [-1,0)$.
Therefore, $|f(x) - f(y)| \leq \delta = \varepsilon$.
A: Here is a non $\epsilon-\delta $ proof: Extend $f$ to a function $g:[0,1]\to \mathbb R$ by setting $g(x)=f(x)$ for all $x\in (0,1)$ and $g(0)=1$, $g(1)=\sin (1)$. Now, $g$ is continuous on its domain which is a closed interval and thus is uniformly continuous. It is very easy to show that the restriction of a uniformly continuous function is itself uniformly continuous and thus $f$ is uniformly continuous. 
Giving a direct $\epsilon -\delta $ proof will be quite technically messy. Is there any particular reason you wish to obtain such a proof. 
