How exactly can't $\delta$ depend on $x$ in the definition of uniform continuity? I'm told that a function defined on an interval $[a,b]$ or $(a,b)$ is uniformly continuous if for each $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that $|x-t|\lt \delta$ implies that $|f(x)-f(t)|\lt \epsilon$. Then it gives a little note saying that $\delta$ cannot depend on $x$, it can only depend on $\epsilon$. 
With ordinary continuity, the $\delta$ can depend on both $x$ and $\epsilon$. I'm just a little lost on why $|x-t|\lt \delta$ implies $|f(x)-f(t)|\lt \epsilon$, and how $\delta$ can't depend on $x$ but only $\epsilon$.

For an trivial example, the function $f(x)=x$ is uniformly continuous on $(0,1)$, because for any $\epsilon>0$, there is a $\delta>0$ such that $|x-t|<\delta$ implies that $|f(x)-f(t)|<\epsilon$; namely, we can just choose $\delta=\epsilon$, and then
$$|x-t|=|f(x)-f(t)|<\epsilon=\delta.$$
In contrast, the function $\sin(\frac{1}{x})$ is continuous, but not uniformly continuous, on the interval $(0,1)$ because, taking (for example) $\epsilon=\frac{1}{2}$, for any $\delta>0$ there are always some $x,t\in (0,1)$ with $|x-t|<\delta$ but $|f(x)-f(t)|\geq \epsilon$; you can just take $x=\frac{1}{2\pi k}$ and $t=\frac{1}{2\pi k+(\pi/2)}$ where $k$ is big enough to ensure that $|x-t|<\delta$.
